Algebra I math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Multiply polynomials

Learning goals

Multiplies a monomial by every term of a polynomial.
Multiplies polynomials by forming all partial products.
Combines like terms in an expanded polynomial product.
Represents a polynomial product with an area model.
Checks a polynomial calculation using its degree and leading term.

Difficulty: 1 – 5

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Multiply polynomials 1

Three identical repair kits each contain \(x\) screws and \(2\) nuts. Write an expression for the total number of pieces in all three kits, first with parentheses and then without parentheses.

Hints

- Find the number of pieces in one kit first. - Each type of piece appears in all three kits.

Solution

1. One kit contains \(x+2\) pieces, so three kits contain \(3(x+2)\). 2. Distributing the \(3\) gives \(3x+6\).

Answer

\(3(x+2)=3x+6\)

Multiply polynomials 2

Expand \(-2(x-3)\). Explain why the constant term in your answer is positive.

Hints

- The outside factor applies to both terms inside parentheses. - Keep the signs with the numbers you multiply.

Solution

1. Distribute \(-2\) to both terms: \((-2)x+(-2)(-3)\). 2. Multiplying two negative numbers gives a positive product, so the result is \(-2x+6\).

Answer

\(-2x+6\); the constant is positive because \((-2)(-3)=6\).

Multiply polynomials 3

Expand \(x(x+6)\). Explain why your result includes an \(x^2\)-term.

Hints

- Think of the expression as a number times a sum. How can you split it into two smaller products? - Test your expansion using \(x=2\). Does it give the same value as the original expression?

Solution

1. Multiply \(x\) by both terms: \(x\cdot x+x\cdot6\). 2. This gives \(x^2+6x\). The squared term comes from multiplying \(x\) by \(x\).

Answer

\(x^2+6x\). The term \(x^2\) comes from \(x\cdot x\).

Multiply polynomials 4

Expand \(-x(2x-3)\). Explain the sign of each term.

Hints

- Treat the negative sign as part of the outside factor. - Find the sign of each product before combining the terms.

Solution

1. The first product is \((-x)(2x)=-2x^2\). 2. The second product is \((-x)(-3)=3x\). 3. Together they give \(-2x^2+3x\), with opposite signs because one product has one negative factor and the other has two.

Answer

\(-2x^2+3x\). A negative times a positive is negative; two negatives multiply to a positive.

Multiply polynomials 5

A first step in multiplying \((x+2)(x+4)\) is to write \(x(x+4)+2(x+4)\). Expand each of these two parts. Do not combine them yet.

Hints

- Work on one of the two written parts at a time. - In each part, multiply the outside term by both terms inside the parentheses.

Solution

1. The first part becomes \(x(x+4)=x^2+4x\). 2. The second becomes \(2(x+4)=2x+8\). 3. These are the two partial products needed to multiply the original binomials.

Answer

First part: \(x^2+4x\) Second part: \(2x+8\)

Multiply polynomials 6

What are the areas of regions A and B? Use the labeled side lengths to explain both products.

Figure for problem 560327

Hints

- Find the width and height of each marked region on the outside edges. - Use length times width for each region separately.

Solution

1. Region A has width \(2\) and height \(x\), so its area is \(2x\). 2. Region B has width \(x\) and height \(1\), so its area is \(x\). 3. These are two of the partial products in the area model for \((x+2)(x+1)\).

Answer

A: \(2x\) B: \(x\)

Multiply polynomials 7

A rectangle has width \(x+5\), where \(x>0\). Its height is split into lengths \(x\) and \(2\). Write the areas of the two parts without expanding. Explain why their sum equals the whole rectangle's area.

Hints

- How could you find the area of each part separately? - Find the whole rectangle's area in another way, without using the separate areas. - Do these two methods count every part of the rectangle exactly once?

Solution

1. The part with height \(x\) has area \(x(x+5)\). 2. The part with height \(2\) has area \(2(x+5)\). 3. The parts fill the rectangle, so \(x(x+5)+2(x+5)=(x+2)(x+5)\).

Answer

The two areas are \(x(x+5)\) and \(2(x+5)\). Their sum is \((x+2)(x+5)\) because together the parts fill the whole rectangle.

Multiply polynomials 8

Rewrite \(f(x)=(2x+1)(x-3)-2x^2+4x\) as a linear expression in the form \(mx+b\).

Hints

- Expand the product before combining any terms. - Group the quadratic terms, linear terms, and constants separately. - Check whether the quadratic terms cancel completely.

Solution

1. Expand the product: \((2x+1)(x-3)=2x^2-5x-3\). 2. Substitute this into the expression: \(f(x)=2x^2-5x-3-2x^2+4x\). 3. Combine like terms. The quadratic terms cancel, and \(-5x+4x=-x\), so \(f(x)=-x-3\).

Answer

\(f(x)=-x-3\)

Multiply polynomials 9

Simplify each expression. a) \(9a - (4a - 7b) - 2b\) b) \(4x(3 - y) + 4xy - 10x\)

Hints

- Remove parentheses before combining terms. - Combine only like terms. - Check the signs carefully, especially when subtracting parentheses.

Solution

1. For a), remove the parentheses and combine like terms: \(9a - 4a + 7b - 2b = 5a + 5b\). 2. For b), distribute and combine like terms: \(4x(3 - y) + 4xy - 10x = 12x - 4xy + 4xy - 10x = 2x\).

Answer

a) \(5a + 5b\) b) \(2x\)

Multiply polynomials 10

Expand each product and simplify. a) \(2x^3(4x^2 - 5x + 1)\) b) \(-ab^2(3a^2b - 2b + 7)\) c) \(4s^2t^3(s^2 - 3st + 2t^2)\)

Hints

- When multiplying powers with the same base, add the exponents. - Pay close attention to the sign when the outside factor is negative. - Multiply the outside monomial by every term in the polynomial.

Solution

1. For a), distribute and add exponents of like bases: \(2x^3(4x^2 - 5x + 1) = 8x^5 - 10x^4 + 2x^3\). 2. For b), \(-ab^2(3a^2b - 2b + 7) = -3a^3b^3 + 2ab^3 - 7ab^2\). 3. For c), \(4s^2t^3(s^2 - 3st + 2t^2) = 4s^4t^3 - 12s^3t^4 + 8s^2t^5\).

Answer

a) \(8x^5 - 10x^4 + 2x^3\) b) \(-3a^3b^3 + 2ab^3 - 7ab^2\) c) \(4s^4t^3 - 12s^3t^4 + 8s^2t^5\)

Multiply polynomials 11

Expand and simplify. 1) \(7x(2x - 3y)\) 2) \(-3a(4b + 2a)\) 3) \(0.4z(10z - 5)\)

Hints

- Multiply the monomial by every term in the parentheses. - Pay close attention to signs when the monomial is negative. - Remember that \(x \cdot x = x^2\).

Solution

1. \(7x(2x - 3y) = 14x^2 - 21xy\). 2. \(-3a(4b + 2a) = -12ab - 6a^2\). 3. \(0.4z(10z - 5) = 4z^2 - 2z\).

Answer

1) \(14x^2 - 21xy\) 2) \(-12ab - 6a^2\) 3) \(4z^2 - 2z\)

Multiply polynomials 12

Use the distributive property to expand. 1) \(9(3x - 5y)\) 2) \(-4(2a + 6b - 3)\) 3) \(2m(4n - 7p)\)

Hints

- Multiply the factor outside the parentheses by every term inside. - Pay close attention to signs when the outside factor is negative. - Multiply both the numerical coefficients and the variable factors.

Solution

1. \(9(3x - 5y) = 27x - 45y\). 2. \(-4(2a + 6b - 3) = -8a - 24b + 12\). 3. \(2m(4n - 7p) = 8mn - 14mp\).

Answer

1) \(27x - 45y\) 2) \(-8a - 24b + 12\) 3) \(8mn - 14mp\)

Multiply polynomials 13

Expand each expression. a) \(5a(2a^2 - 3a + 4)\) b) \(-2x^2(3x - 5)\) c) \((4y^2 - y - 6)(-3y)\)

Hints

- Multiply the monomial by every term in the polynomial. - Pay close attention to signs when the monomial is negative. - When multiplying powers with the same base, add their exponents.

Solution

1. For a), \(5a(2a^2 - 3a + 4) = 10a^3 - 15a^2 + 20a\). 2. For b), \(-2x^2(3x - 5) = -6x^3 + 10x^2\). 3. For c), \((4y^2 - y - 6)(-3y) = -12y^3 + 3y^2 + 18y\).

Answer

a) \(10a^3 - 15a^2 + 20a\) b) \(-6x^3 + 10x^2\) c) \(-12y^3 + 3y^2 + 18y\)

Multiply polynomials 14

Expand each expression. 1) \(5x(3x^2 - 2xy + 4y^2)\) 2) \(-4ab(2a - 5b - 1)\) 3) \(\frac{1}{2}m(4m^2 + 8mn - 6n^2)\)

Hints

- Multiply the outside monomial by every term inside the parentheses. - Pay close attention to signs when multiplying by a negative monomial. - Add exponents when multiplying powers with the same base. - Writing each partial product can prevent errors.

Solution

1. \(5x(3x^2 - 2xy + 4y^2) = 15x^3 - 10x^2y + 20xy^2\). 2. \(-4ab(2a - 5b - 1) = -8a^2b + 20ab^2 + 4ab\). 3. \(\frac{1}{2}m(4m^2 + 8mn - 6n^2) = 2m^3 + 4m^2n - 3mn^2\).

Answer

1) \(15x^3 - 10x^2y + 20xy^2\) 2) \(-8a^2b + 20ab^2 + 4ab\) 3) \(2m^3 + 4m^2n - 3mn^2\)

Multiply polynomials 15

Use the distributive property to expand. 1) \(6(3a - 4b + 7)\) 2) \((x^2 - 2x - 5)(-3)\) 3) \(0.4(10m - 5n + 15)\) 4) \((9r + 6s - 3)\left(-\frac{2}{3}\right)\)

Hints

- Multiply the outside factor by every term inside the parentheses. - Check the sign of each product, especially when two negative factors are multiplied. - Fractions and decimals distribute in the same way as integers.

Solution

1. \(6(3a - 4b + 7) = 18a - 24b + 42\). 2. \((x^2 - 2x - 5)(-3) = -3x^2 + 6x + 15\). 3. \(0.4(10m - 5n + 15) = 4m - 2n + 6\). 4. \((9r + 6s - 3)\left(-\frac{2}{3}\right) = -6r - 4s + 2\).

Answer

1) \(18a - 24b + 42\) 2) \(-3x^2 + 6x + 15\) 3) \(4m - 2n + 6\) 4) \(-6r - 4s + 2\)

Multiply polynomials 16

Expand and simplify. 1) \(5x(3x^2 - 4x + 2)\) 2) \(-2ab(4a^2 - 3ab + b^2)\)

Hints

- Multiply the outside monomial by every term in the polynomial. - Pay close attention to signs when multiplying by a negative monomial. - Add exponents when multiplying powers with the same base.

Solution

1. \(5x(3x^2 - 4x + 2) = 15x^3 - 20x^2 + 10x\). 2. \(-2ab(4a^2 - 3ab + b^2) = -8a^3b + 6a^2b^2 - 2ab^3\).

Answer

1) \(15x^3 - 20x^2 + 10x\) 2) \(-8a^3b + 6a^2b^2 - 2ab^3\)

Multiply polynomials 17

Expand and simplify. \(-1.5a(4a^2 - 2a + 0.8)\)

Hints

- Multiply the monomial by every term in the polynomial. - Pay close attention to signs when multiplying by a negative number. - Add exponents when multiplying powers with the same base. - Break the work into three separate products.

Solution

1. Distribute \(-1.5a\): \(-1.5a(4a^2 - 2a + 0.8) = -6a^3 + 3a^2 - 1.2a\).

Answer

\(-6a^3 + 3a^2 - 1.2a\)

Multiply polynomials 18

Expand and simplify. 1) \((x + 7)(x - 2)\) 2) \((3a - 4)(2a + 5)\) 3) \((5 - 2b)(b + 3)\)

Hints

- Multiply every term in the first binomial by every term in the second. - Check signs carefully. - Combine terms only when their variable parts and exponents match.

Solution

1. \((x + 7)(x - 2) = x^2 - 2x + 7x - 14 = x^2 + 5x - 14\). 2. \((3a - 4)(2a + 5) = 6a^2 + 15a - 8a - 20 = 6a^2 + 7a - 20\). 3. \((5 - 2b)(b + 3) = 5b + 15 - 2b^2 - 6b = -2b^2 - b + 15\).

Answer

1) \(x^2 + 5x - 14\) 2) \(6a^2 + 7a - 20\) 3) \(-2b^2 - b + 15\)

Multiply polynomials 19

Expand and simplify. 1) \((x + 8)(x - 2)\) 2) \((a - 4)(b + 5)\) 3) \((y - 3)(y - 6)\)

Hints

- Multiply every term in the first binomial by every term in the second. - Pay close attention to the signs of products involving negative terms. - Combine like terms after expanding.

Solution

1. \((x + 8)(x - 2) = x^2 - 2x + 8x - 16 = x^2 + 6x - 16\). 2. \((a - 4)(b + 5) = ab + 5a - 4b - 20\). 3. \((y - 3)(y - 6) = y^2 - 6y - 3y + 18 = y^2 - 9y + 18\).

Answer

1) \(x^2 + 6x - 16\) 2) \(ab + 5a - 4b - 20\) 3) \(y^2 - 9y + 18\)

Multiply polynomials 20

Expand and simplify. 1) \((4x + 3)(2x - 5)\) 2) \((a - 6b)(2a + b)\) 3) \((y^2 + 3)(y^2 - 4)\)

Hints

- Multiply every term in one factor by every term in the other. - Apply the sign rules carefully. - Combine terms only when their variables and exponents match exactly.

Solution

1. \((4x + 3)(2x - 5) = 8x^2 - 20x + 6x - 15 = 8x^2 - 14x - 15\). 2. \((a - 6b)(2a + b) = 2a^2 + ab - 12ab - 6b^2 = 2a^2 - 11ab - 6b^2\). 3. \((y^2 + 3)(y^2 - 4) = y^4 - 4y^2 + 3y^2 - 12 = y^4 - y^2 - 12\).

Answer

1) \(8x^2 - 14x - 15\) 2) \(2a^2 - 11ab - 6b^2\) 3) \(y^4 - y^2 - 12\)

Multiply polynomials 21

Expand and simplify. 1) \((a + 6)(3a - 2)\) 2) \(4z + 2 \cdot 3(2z - 5)\)

Hints

- In a product of two binomials, multiply every term in one factor by every term in the other. - In a chain of multiplication, factors may be regrouped and reordered. - Combine only terms with matching variables and exponents. - Track all signs carefully.

Solution

1. \((a + 6)(3a - 2) = 3a^2 - 2a + 18a - 12 = 3a^2 + 16a - 12\). 2. Use the associative property: \(2 \cdot 3(2z - 5) = 6(2z - 5) = 12z - 30\). Then \(4z + 12z - 30 = 16z - 30\).

Answer

1) \(3a^2 + 16a - 12\) 2) \(16z - 30\)

Multiply polynomials 22

Expand both products in \((n+3)(n+4)-(n+1)(n+6)\), simplify the difference, and explain why the final value does not depend on \(n\).

Hints

- Expand each product separately. - Subtract the entire second polynomial. - Identify which variable terms cancel and what remains.

Solution

1. \((n+3)(n+4)=n^2+7n+12\). 2. \((n+1)(n+6)=n^2+7n+6\). 3. Subtract: \((n^2+7n+12)-(n^2+7n+6)=6\). 4. The \(n^2\) and \(n\) terms cancel completely, so the result is the constant \(6\) for every value of \(n\).

Answer

First product: \(n^2+7n+12\) Second product: \(n^2+7n+6\) Difference: \(6\) The value is independent of \(n\) because all variable terms cancel.

Multiply polynomials 23

Write the rectangle's area as a product of its full side lengths and as a sum of the four smaller areas. Then expand the product.

Figure for problem 554968

Hints

- Find each full side length from its two labeled pieces. - Work out the areas of the four smaller rectangles separately. - Compare the area of the whole rectangle with the sum of its parts.

Solution

1. The full side lengths are \(x+4\) and \(2x+1\), giving area \((x+4)(2x+1)\). 2. The four parts have areas \(2x^2\), \(8x\), \(x\), and \(4\). 3. Their sum is \(2x^2+8x+x+4=2x^2+9x+4\). 4. Therefore, \((x+4)(2x+1)=2x^2+9x+4\).

Answer

Product: \((x+4)(2x+1)\) Sum of four parts: \(2x^2+8x+x+4\) Expanded: \(2x^2+9x+4\)

Multiply polynomials 24

Use the side labels to write the area of the whole rectangle as a product. Then write it as the sum of the two smaller areas. Explain why the two expressions are equal.

Figure for problem 560311

Hints

- Find the total width by combining the bottom segments. - Calculate each smaller rectangle's area separately. - What region does each calculation measure?

Solution

1. The full rectangle has height \(x\) and width \(x+4\), so its area is \(x(x+4)\). 2. The two smaller rectangles have areas \(x\cdot x=x^2\) and \(x\cdot4=4x\). 3. Because the parts fill the whole rectangle, \(x(x+4)=x^2+4x\).

Answer

The whole area is \(x(x+4)\); the two parts sum to \(x^2+4x\), so the expressions are equal.

Multiply polynomials 25

Expand and simplify \(x(2x+5)\). State the degree of the result.

Hints

- Multiply the outside variable by each term separately. - When multiplying equal variable bases, add the exponents. - Look for the greatest exponent in the simplified result.

Solution

1. Multiply \(x\) by both terms: \(x\cdot2x+x\cdot5\). 2. This gives \(2x^2+5x\), whose highest power is \(2\).

Answer

\(2x^2+5x\); degree \(2\).

Multiply polynomials 26

Expand \((x+2)(x+3)\). First fill in the four products: \(x\cdot x=\square\), \(x\cdot3=\square\), \(2\cdot x=\square\), \(2\cdot3=\square\). Then combine the terms.

Hints

- Each term in the first binomial must multiply each term in the second. - Keep the four products separate until all are found. - Then look for terms with the same variable power.

Solution

1. The four products are \(x^2\), \(3x\), \(2x\), and \(6\). 2. Adding them gives \(x^2+3x+2x+6=x^2+5x+6\).

Answer

Blanks: \(x^2\), \(3x\), \(2x\), \(6\). Expanded result: \(x^2+5x+6\).

Multiply polynomials 27

Complete the two partial products, then combine them to multiply \((x+1)(x^2+2)\): \(x(x^2+2)=\square\) \(1(x^2+2)=\square\).

Hints

- Treat each term of the first factor as a multiplier for the whole second factor. - Keep track of the powers when multiplying by \(x\). - Add the two partial results only after expanding them.

Solution

1. Multiplying by \(x\) gives \(x^3+2x\). 2. Multiplying by \(1\) gives \(x^2+2\). 3. Their sum is \(x^3+x^2+2x+2\).

Answer

First blank: \(x^3+2x\). Second blank: \(x^2+2\). Product: \(x^3+x^2+2x+2\).

Multiply polynomials 28

Expand \((x+y)(x+2y)\). Write the four partial products before combining like terms.

Hints

- Multiply each term in the first factor by each term in the second. - Remember that \(xy\) and \(yx\) describe the same product. - Combine only the genuinely like terms.

Solution

1. The four products are \(x^2\), \(2xy\), \(xy\), and \(2y^2\). 2. Since \(2xy+xy=3xy\), the expanded expression is \(x^2+3xy+2y^2\).

Answer

Partial products: \(x^2\), \(2xy\), \(xy\), \(2y^2\). Simplified: \(x^2+3xy+2y^2\).

Multiply polynomials 29

Maya and Luis calculate the rectangle's area: - Maya: \(2x^2+3x\) - Luis: \(2x^2+6x\) Who is correct? Use the two parts of the diagram to explain.

Figure for problem 560320

Hints

- Both small rectangles share the same full height. - Find each small area before choosing a student's answer. - Compare the second small area's calculation with the proposed terms.

Solution

1. The left section has height \(2x\) and width \(x\), so its area is \(2x^2\). 2. The right section has height \(2x\) and width \(3\), so its area is \(6x\). 3. The total is \(2x^2+6x\), so Luis is correct. Maya did not multiply the second width by the full height.

Answer

Luis: \(2x^2+6x\). The parts have areas \(2x^2\) and \(6x\).

Multiply polynomials 30

The strip marked \(R\) is removed from the rectangle. Write the remaining area as the whole area minus the removed strip's area. Then simplify.

Figure for problem 560321

Hints

- Read the two horizontal pieces to find the original width. - A removed strip contributes a product of its width and the full height. - Subtract the removed area before simplifying the polynomial.

Solution

1. The full width is \((x+3)+2=x+5\) and the height is \(x\), so the whole area is \(x(x+5)\). 2. The removed strip has width \(2\) and height \(x\), so its area is \(2x\). 3. The remaining area is \(x(x+5)-2x=x^2+5x-2x=x^2+3x\).

Answer

\(x(x+5)-2x=x^2+3x\)

Multiply polynomials 31

A student multiplies \((2x+1)(x+3)\) and gets \(2x^2+3\). Explain the mistake and find the correct product.

Hints

- Test the student's expression and the original product at an easy value of \(x\). - If they differ, expand the original product to find out why. - Which parts of your calculation explain the mismatch?

Solution

1. The student multiplied the first terms and the last terms but omitted the two mixed products. 2. All four products are \(2x^2\), \(6x\), \(x\), and \(3\). 3. Adding and combining gives \(2x^2+7x+3\).

Answer

Correct product: \(2x^2+7x+3\). The missing mixed products total \(7x\).

Multiply polynomials 32

Expand \((x-1)(x+2)\) without a list of partial products. Show how you found the terms before combining them.

Hints

- Think about how every term in one pair of parentheses affects the other pair. - Keep the negative sign with the second term of the first factor. - Check by substituting a small value for \(x\) into both expressions.

Solution

1. Distribute both terms of the first factor: \(x(x+2)-1(x+2)\). 2. Expanding gives \(x^2+2x-x-2\). 3. Combining like terms gives \(x^2+x-2\).

Answer

\((x-1)(x+2)=x^2+2x-x-2=x^2+x-2\)

Multiply polynomials 33

A square flower bed and a rectangular vegetable bed have the same area. The rectangle is \(4\,\text{ft}\) longer than the side of the square and \(3\,\text{ft}\) narrower than the side of the square. Let \(s\) be the side length of the square. a) Write the square's area and the rectangle's area as polynomial expressions in \(s\). Expand the rectangle's area. b) Set the two area expressions equal and find \(s\).

Hints

- Express each rectangle dimension in terms of the square's side length. - Multiply the two rectangle dimensions before comparing areas. - Look for cancellation after setting the area expressions equal.

Solution

1. The square's area is \(s^2\). 2. The rectangle has dimensions \(s+4\) and \(s-3\), so its area is \((s+4)(s-3)=s^2+s-12\). 3. Equal areas give \(s^2=s^2+s-12\). 4. Subtract \(s^2\) from both sides: \(s-12=0\), so \(s=12\). 5. The square's side length is \(12\,\text{ft}\).

Answer

a) Square area: \(s^2\) Rectangle area: \((s+4)(s-3)=s^2+s-12\) b) \(s^2=s^2+s-12\), so \(s=12\,\text{ft}\)

Multiply polynomials 34

A rectangular sign is twice as long as it is tall. If its height is increased by \(3\,\text{in.}\) and its length is decreased by \(2\,\text{in.}\), its area increases by \(30\,\text{in.}^2\). Let \(x\) be the original height. a) Write and expand the original and new area expressions in \(x\). b) Use the \(30\,\text{in.}^2\) area increase to find the sign's original height and length.

Hints

- Express the original length in terms of the height. - Apply both dimension changes before writing the new area. - Expand the new area before using the stated area increase.

Solution

1. The original height is \(x\) and the original length is \(2x\), so the original area is \(2x^2\). 2. The new height is \(x+3\), and the new length is \(2x-2\). 3. The new area is \((x+3)(2x-2)=2x^2+4x-6\). 4. The new area is \(30\,\text{in.}^2\) greater, so \(2x^2+4x-6=2x^2+30\). 5. Simplify: \(4x-6=30\), so \(x=9\). 6. The original dimensions were \(9\,\text{in.}\) by \(18\,\text{in.}\).

Answer

a) Original area: \(2x^2\) New area: \((x+3)(2x-2)=2x^2+4x-6\) b) \(2x^2+4x-6=2x^2+30\), so the original dimensions were \(9\,\text{in.}\) by \(18\,\text{in.}\).

Multiply polynomials 35

Consider \(f(x)=\frac{1}{2}x(4-2x)+x^2+3\) and \(g(x)=2(x+1.5)\). Rewrite both expressions in simplified linear form and determine whether they are equivalent for every value of \(x\).

Hints

- Simplify each expression separately. - In \(f\), check what happens to the two \(x^2\)-terms. - Equivalent expressions have the same simplified form for every value of the variable.

Solution

1. Distribute in \(f\): \(\frac{1}{2}x(4-2x)=2x-x^2\), so \(f(x)=2x-x^2+x^2+3=2x+3\). 2. Distribute in \(g\): \(g(x)=2x+3\). 3. Both expressions simplify to \(2x+3\), so they are equivalent for every value of \(x\).

Answer

\(f(x)=2x+3\) and \(g(x)=2x+3\). Yes, they are equivalent for every value of \(x\).

Multiply polynomials 36

Simplify each expression. a) \(\frac{2}{5}m + \frac{3}{4}n - \left(\frac{1}{10}m - \frac{1}{2}n\right)\) b) \(4a(3b) - 2a(b + 5) + ab\) c) \(x^2 - \left[4x - (2x^2 + x)\right] + 7\)

Hints

- Use common denominators before adding or subtracting fractional coefficients. - With nested grouping symbols, work from the inside out. - Products such as \(ab\) are like terms only with other terms having exactly the same variable factors.

Solution

1. For a), remove the parentheses and combine like terms: \(\left(\frac{2}{5} - \frac{1}{10}\right)m + \left(\frac{3}{4} + \frac{1}{2}\right)n = \frac{3}{10}m + \frac{5}{4}n\). 2. For b), expand and combine like terms: \(12ab - 2ab - 10a + ab = 11ab - 10a\). 3. For c), work from the inside out: \(x^2 - (4x - 2x^2 - x) + 7 = x^2 - 3x + 2x^2 + 7 = 3x^2 - 3x + 7\).

Answer

a) \(\frac{3}{10}m + \frac{5}{4}n\) b) \(11ab - 10a\) c) \(3x^2 - 3x + 7\)

Multiply polynomials 37

One side of a rectangle is exactly \(5\,\text{in.}\) longer than the other side. If the longer side is increased by \(2\,\text{in.}\) and the shorter side is increased by \(3\,\text{in.}\), the area increases by \(66\,\text{in.}^2\). Let \(x\) be the original shorter side. a) Write and expand the original and new area expressions. b) Subtract the original area from the new area, then use the \(66\,\text{in.}^2\) increase to find the original side lengths.

Hints

- Express both original sides using one variable. - Apply the two different increases to the correct sides. - Expand both area products and subtract them before solving.

Solution

1. The original sides are \(x\) and \(x+5\), so the original area is \(x(x+5)=x^2+5x\). 2. The new sides are \(x+3\) and \(x+7\), so the new area is \((x+3)(x+7)=x^2+10x+21\). 3. The area increase is \((x^2+10x+21)-(x^2+5x)=5x+21\). 4. Set \(5x+21=66\). Then \(5x=45\), so \(x=9\). 5. The original side lengths were \(9\,\text{in.}\) and \(14\,\text{in.}\).

Answer

a) Original area: \(x^2+5x\) New area: \(x^2+10x+21\) b) Area increase: \(5x+21\) \(5x+21=66\), so the original side lengths were \(9\,\text{in.}\) and \(14\,\text{in.}\).

Multiply polynomials 38

Simplify each expression completely. Pay attention to exponents and to equivalent variable products. a) \(4x^2y-2xy^2+3yx^2-5y^2x\) b) \(\frac{3}{4}ab-\frac{1}{2}ba+2.5ab\) c) \(z(2z-3)-(z^2+4z)\)

Hints

- Products such as \(x^2y\) and \(yx^2\) have the same variable part. - Express fractions and decimals in a common form before combining coefficients. - Distribute carefully in part c, including the negative sign before the second set of parentheses.

Solution

1. For part a, use the commutative property to recognize \(yx^2=x^2y\) and \(y^2x=xy^2\). Then combine like terms: \(4x^2y+3x^2y-2xy^2-5xy^2=7x^2y-7xy^2\). 2. For part b, \(ba=ab\). Combine the coefficients: \(\frac{3}{4}-\frac{1}{2}+2.5=0.75-0.5+2.5=2.75\). The result is \(2.75ab\), or \(\frac{11}{4}ab\). 3. For part c, distribute and remove the parentheses: \(z(2z-3)-(z^2+4z)=2z^2-3z-z^2-4z=z^2-7z\).

Answer

a) \(7x^2y-7xy^2\) b) \(2.75ab\), or \(\frac{11}{4}ab\) c) \(z^2-7z\)

Multiply polynomials 39

Fill in each blank so the equation is true. a) \(\square(3x-4y)=12x^2-16xy\) b) \(-5b(2a+\square)=-10ab-15b^2\) c) \(x(\square-7)=x^2-7x\)

Hints

- Compare corresponding product terms to determine what the missing factor must contribute. - Check both signs and variable powers. - Verify each completed product by expanding it.

Solution

1. Compare the first product term with \(3x\): \(12x^2\div3x=4x\). Checking gives \(4x(-4y)=-16xy\). 2. Compare the second product term with \(-5b\): \(-15b^2\div(-5b)=3b\). 3. Compare \(x^2\) with the outside factor \(x\): \(x^2\div x=x\).

Answer

a) \(4x\) b) \(3b\) c) \(x\)

Multiply polynomials 40

Expand and simplify. a) \(\frac{2}{3}z(6z^2 - 9z + 12)\) b) \(-1.5m(2m^2 - 4m + 6)\) c) \((10x^3 - 5x^2 + 25x)\left(-\frac{1}{5}x\right)\)

Hints

- Determine the sign of each product before simplifying it. - Fractions and decimals follow the same multiplication rules as integers. - When multiplying powers with the same base, add the exponents. A variable with no written exponent has exponent \(1\).

Solution

1. For a), \(\frac{2}{3}z(6z^2 - 9z + 12) = 4z^3 - 6z^2 + 8z\). 2. For b), \(-1.5m(2m^2 - 4m + 6) = -3m^3 + 6m^2 - 9m\). 3. For c), \((10x^3 - 5x^2 + 25x)\left(-\frac{1}{5}x\right) = -2x^4 + x^3 - 5x^2\).

Answer

a) \(4z^3 - 6z^2 + 8z\) b) \(-3m^3 + 6m^2 - 9m\) c) \(-2x^4 + x^3 - 5x^2\)

Multiply polynomials 41

Lucas and Sarah expand \(-3x^2(2x - 4y + 1)\). Lucas gets \(-6x^3 + 12x^2y - 1\). Sarah gets \(-6x^3 + 12x^2y - 3x^2\). Determine who is correct, and briefly describe the other student's error.

Hints

- Expand the expression yourself before comparing the answers. - Was every term inside the parentheses multiplied by the outside monomial? - What is the product of a quantity and \(1\)?

Solution

1. Distribute \(-3x^2\) to every term: \(-3x^2(2x - 4y + 1) = -6x^3 + 12x^2y - 3x^2\). 2. Sarah is correct. 3. Lucas changed \(+1\) to \(-1\) instead of multiplying it by \(-3x^2\).

Answer

Sarah is correct. Lucas failed to multiply the final term \(1\) by \(-3x^2\).

Multiply polynomials 42

Expand and simplify. \(5x(x - 2y) - 3y(4x - y) + 22xy\)

Hints

- Expand one product at a time. - Pay close attention to the negative factor before the second product. - Combine terms only when they have the same variable factors. - Check that every product was expanded completely.

Solution

1. Expand the products: \(5x^2 - 10xy - 12xy + 3y^2 + 22xy\). 2. Combine the \(xy\)-terms: \((-10 - 12 + 22)xy = 0\). 3. The simplified expression is \(5x^2 + 3y^2\).

Answer

\(5x^2 + 3y^2\)

Multiply polynomials 43

Expand and simplify. \(0.5k(4k - 6m) - 2m(k - 3m) - 2k^2\)

Hints

- Distribute the negative factor carefully. - Remember that \(km\) and \(mk\) are the same product. - Look for terms that cancel. - Multiply the decimal coefficient just as you would any other number.

Solution

1. Expand the first product: \(0.5k(4k - 6m) = 2k^2 - 3km\). 2. Expand the second product: \(-2m(k - 3m) = -2km + 6m^2\). 3. Combine all terms: \(2k^2 - 3km - 2km + 6m^2 - 2k^2 = 6m^2 - 5km\).

Answer

\(6m^2 - 5km\)

Multiply polynomials 44

Find the product. \(5xy\left(\frac{2}{5}x^2y - 10xy^2 + \frac{1}{2}y\right)\)

Hints

- Multiply every term in the polynomial by the monomial outside. - Multiply the coefficients first, then handle each variable. - When multiplying \(x^2\) by \(x\), add the exponents. - Choose either fraction or decimal form for the coefficient calculation and use it consistently.

Solution

1. Multiply the first term: \(5xy\left(\frac{2}{5}x^2y\right) = 2x^3y^2\). 2. Multiply the second term: \(5xy(-10xy^2) = -50x^2y^3\). 3. Multiply the third term: \(5xy\left(\frac{1}{2}y\right) = \frac{5}{2}xy^2\). 4. The product is \(2x^3y^2 - 50x^2y^3 + \frac{5}{2}xy^2\).

Answer

\(2x^3y^2 - 50x^2y^3 + \frac{5}{2}xy^2\)

Multiply polynomials 45

Expand and simplify. \(4x(3x - 2y) - 2x(y + 5x) + 7xy\)

Hints

- Pay close attention to the sign of each outside factor. - Group terms by their variable factors before combining. - Terms can be combined only when their variables and exponents match exactly.

Solution

1. Expand both products: \(12x^2 - 8xy - 2xy - 10x^2 + 7xy\). 2. Combine the \(x^2\)-terms: \(12x^2 - 10x^2 = 2x^2\). 3. Combine the \(xy\)-terms: \(-8xy - 2xy + 7xy = -3xy\). 4. The result is \(2x^2 - 3xy\).

Answer

\(2x^2 - 3xy\)

Multiply polynomials 46

Expand and simplify. \(2a(4b - 3a) - 5(a^2 - 2ab) - a(3a + b)\)

Hints

- Distribute negative factors carefully. - Expand each product separately before combining terms. - Group terms with the same variable factors.

Solution

1. Expand each product: \(8ab - 6a^2 - 5a^2 + 10ab - 3a^2 - ab\). 2. Combine the \(a^2\)-terms: \(-6a^2 - 5a^2 - 3a^2 = -14a^2\). 3. Combine the \(ab\)-terms: \(8ab + 10ab - ab = 17ab\). 4. The result is \(-14a^2 + 17ab\).

Answer

\(-14a^2 + 17ab\)

Multiply polynomials 47

Expand and simplify. \(1.2a(2a - 0.5b) - 0.4(3a^2 - 5ab + 2b^2) + 0.5b(4a - 6b)\)

Hints

- Multiply every term inside each set of parentheses by its outside factor. - Check the sign of each product. - Group the \(a^2\)-terms, \(ab\)-terms, and \(b^2\)-terms separately. - Remember that \(ab = ba\).

Solution

1. Expand the first product: \(1.2a(2a - 0.5b) = 2.4a^2 - 0.6ab\). 2. Expand the second product: \(-0.4(3a^2 - 5ab + 2b^2) = -1.2a^2 + 2ab - 0.8b^2\). 3. Expand the third product: \(0.5b(4a - 6b) = 2ab - 3b^2\). 4. Combine like terms: \(1.2a^2 + 3.4ab - 3.8b^2\).

Answer

\(1.2a^2 + 3.4ab - 3.8b^2\)

Multiply polynomials 48

Expand and simplify. \(1.5(4x^2 - 2x + 6) - 1.2(5x^2 - 10x - 5)\)

Hints

- Distribute each decimal coefficient to every term. - Pay close attention to the minus sign before the second product. - Do not combine \(x^2\)-terms with \(x\)-terms. - Remove all parentheses before grouping like terms.

Solution

1. Expand the first product: \(1.5(4x^2 - 2x + 6) = 6x^2 - 3x + 9\). 2. Expand the second product and subtract it: \(-1.2(5x^2 - 10x - 5) = -6x^2 + 12x + 6\). 3. Combine like terms: \(6x^2 - 3x + 9 - 6x^2 + 12x + 6 = 9x + 15\).

Answer

\(9x + 15\)

Multiply polynomials 49

A rectangle is three times as long as it is wide. When its length is shortened by \(2\,\text{in.}\), its area decreases by \(12\,\text{in.}^2\). a) Let \(w\) be the original width. Write and expand polynomial expressions for the original area and the shortened rectangle's area. Subtract them to obtain an expression for the area decrease. b) Use that expression and the \(12\,\text{in.}^2\) decrease to find the original width and length. c) Explain geometrically why your simplified area-decrease expression makes sense.

Hints

- Express the original length in terms of \(w\). - Expand the new area before subtracting. - Simplify the difference of the two area polynomials. - Compare the algebraic difference with the dimensions of the removed strip.

Solution

1. The original length is \(3w\), so the original area is \(3w^2\). 2. The shortened length is \(3w-2\), so the new area is \(w(3w-2)=3w^2-2w\). 3. The area decrease is \(3w^2-(3w^2-2w)=2w\). 4. Set \(2w=12\). Then \(w=6\), so the original length is \(18\,\text{in.}\). 5. Geometrically, shortening the length by \(2\,\text{in.}\) removes a \(2\,\text{in.}\)-by-\(w\) strip, whose area is \(2w\).

Answer

a) Original area: \(3w^2\); shortened area: \(3w^2-2w\); area decrease: \(2w\) b) Width: \(6\,\text{in.}\); length: \(18\,\text{in.}\) c) The removed strip has area \(2w\), matching the simplified polynomial difference.

Multiply polynomials 50

Elena and Sam each start with a square garden bed of side length \(a\) feet. Elena extends one side by \(2\,\text{ft}\), while Sam extends one side by \(5\,\text{ft}\). Sam's new bed has \(15\,\text{ft}^2\) more area than Elena's. Write and expand the area expression for each new bed. Subtract the expanded expressions and use the difference to find \(a\).

Hints

- Use length times width for each new rectangular bed. - Expand both areas before subtracting them. - Which terms cancel when the larger area is compared with the smaller one? - Use the given area difference after finding its expression.

Solution

1. Elena's new area is \(a(a+2)=a^2+2a\). 2. Sam's new area is \(a(a+5)=a^2+5a\). 3. Their difference is \((a^2+5a)-(a^2+2a)=3a\). 4. The difference is \(15\), so \(3a=15\) and \(a=5\,\text{ft}\).

Answer

Elena: \(a(a+2)=a^2+2a\) Sam: \(a(a+5)=a^2+5a\) Difference: \(3a\) Original side length: \(a=5\,\text{ft}\)

Multiply polynomials 51

Expand and simplify. 1) \((4a + b)(a - 3b)\) 2) \((2x - 3y)(5x - 2y)\) 3) \((m + n)(m - n)\) 4) \((x + 2)(x + 3) - x(x + 5)\)

Hints

- Remember that \(ab = ba\) when identifying like terms. - In part 4, expand both products before subtracting. - Look for terms that cancel in part 3.

Solution

1. \((4a + b)(a - 3b) = 4a^2 - 12ab + ab - 3b^2 = 4a^2 - 11ab - 3b^2\). 2. \((2x - 3y)(5x - 2y) = 10x^2 - 4xy - 15xy + 6y^2 = 10x^2 - 19xy + 6y^2\). 3. \((m + n)(m - n) = m^2 - mn + mn - n^2 = m^2 - n^2\). 4. \((x + 2)(x + 3) - x(x + 5) = x^2 + 5x + 6 - x^2 - 5x = 6\).

Answer

1) \(4a^2 - 11ab - 3b^2\) 2) \(10x^2 - 19xy + 6y^2\) 3) \(m^2 - n^2\) 4) \(6\)

Multiply polynomials 52

Expand and simplify. a) \((y + 4)(y^2 - 4y + 16)\) b) \((z - 2)(z^2 + 2z + 4)\)

Hints

- Multiply every term in the first factor by every term in the second. - Check the signs of all partial products. - After expanding, look for pairs of terms that cancel.

Solution

1. For a), \((y + 4)(y^2 - 4y + 16) = y^3 - 4y^2 + 16y + 4y^2 - 16y + 64 = y^3 + 64\). 2. For b), \((z - 2)(z^2 + 2z + 4) = z^3 + 2z^2 + 4z - 2z^2 - 4z - 8 = z^3 - 8\).

Answer

a) \(y^3 + 64\) b) \(z^3 - 8\)

Multiply polynomials 53

Expand the product, then simplify the entire expression. \((2x + 1)(4x^2 - 2x + 1) - 8x^3\)

Hints

- Expand the product completely before considering the final subtraction. - Remember that \(x \cdot x^2 = x^3\). - Look for terms with equal magnitude and opposite signs.

Solution

1. Expand the product: \((2x + 1)(4x^2 - 2x + 1) = 8x^3 - 4x^2 + 2x + 4x^2 - 2x + 1\). 2. Combine terms in the product: \(8x^3 + 1\). 3. Subtract \(8x^3\): \(8x^3 + 1 - 8x^3 = 1\).

Answer

\(1\)

Multiply polynomials 54

Expand each product and write the result in descending powers of \(x\), \(a\), or \(k\), respectively. 1) \((2x+1)(x-4)\) 2) \((3a-b)(2a+b)\) 3) \((5-k)(k+2)\)

Hints

- Multiply each term in one factor by each term in the other. - Look for terms that have the same variables to the same powers. - Order the terms by the stated variable after simplifying.

Solution

1. \((2x+1)(x-4)=2x^2-8x+x-4=2x^2-7x-4\). 2. \((3a-b)(2a+b)=6a^2+3ab-2ab-b^2=6a^2+ab-b^2\). 3. \((5-k)(k+2)=5k+10-k^2-2k=-k^2+3k+10\).

Answer

1) \(2x^2-7x-4\) 2) \(6a^2+ab-b^2\) 3) \(-k^2+3k+10\)

Multiply polynomials 55

Expand and combine like terms. 1) \((2x^2 - 3y^2)(4x^2 + y^2)\) 2) \((3a^2b + 2a)(5ab^2 - 4b)\) 3) \((-x^2y - 2)(3x^2y + 5)\)

Hints

- Add exponents when multiplying powers with the same base. - Track every negative sign through each partial product. - Like terms must match in every variable and exponent.

Solution

1. \((2x^2 - 3y^2)(4x^2 + y^2) = 8x^4 + 2x^2y^2 - 12x^2y^2 - 3y^4 = 8x^4 - 10x^2y^2 - 3y^4\). 2. \((3a^2b + 2a)(5ab^2 - 4b) = 15a^3b^3 - 12a^2b^2 + 10a^2b^2 - 8ab = 15a^3b^3 - 2a^2b^2 - 8ab\). 3. \((-x^2y - 2)(3x^2y + 5) = -3x^4y^2 - 5x^2y - 6x^2y - 10 = -3x^4y^2 - 11x^2y - 10\).

Answer

1) \(8x^4 - 10x^2y^2 - 3y^4\) 2) \(15a^3b^3 - 2a^2b^2 - 8ab\) 3) \(-3x^4y^2 - 11x^2y - 10\)

Multiply polynomials 56

Expand and simplify. \((a - 2b)(a^2 + 2ab + 4b^2 - 1)\)

Hints

- Multiply the first term of the left factor by the entire second factor, then do the same with the second term. - Use the power rules, such as \(a \cdot a^2 = a^3\). - Pay close attention when multiplying by \(-2b\). - Group matching variable terms to find cancellations.

Solution

1. Multiply \(a\) by every term: \(a^3 + 2a^2b + 4ab^2 - a\). 2. Multiply \(-2b\) by every term: \(-2a^2b - 4ab^2 - 8b^3 + 2b\). 3. Combine like terms. The \(a^2b\)-terms and \(ab^2\)-terms cancel, leaving \(a^3 - 8b^3 - a + 2b\).

Answer

\(a^3 - 8b^3 - a + 2b\)

Multiply polynomials 57

Expand and simplify. \((x^2 - 4xy + 3y^2)(x + 2y)\)

Hints

- There should be six partial products before combining terms. - Check each sign carefully. - Combine only terms with exactly the same variables and exponents. - Mark each term after using it to avoid omissions.

Solution

1. Form all six partial products: \(x^3 + 2x^2y - 4x^2y - 8xy^2 + 3xy^2 + 6y^3\). 2. Combine like terms: \(x^3 + (2 - 4)x^2y + (-8 + 3)xy^2 + 6y^3\). 3. The result is \(x^3 - 2x^2y - 5xy^2 + 6y^3\).

Answer

\(x^3 - 2x^2y - 5xy^2 + 6y^3\)

Multiply polynomials 58

Expand and simplify. \((2a - 3b)(3a^2 + ab - 2b^2)\)

Hints

- Remember that a negative times a negative is positive. - Arrange like terms next to each other before combining. - Check that every term in one factor was multiplied by every term in the other.

Solution

1. Form all partial products: \(6a^3 + 2a^2b - 4ab^2 - 9a^2b - 3ab^2 + 6b^3\). 2. Combine like terms: \(6a^3 + (2 - 9)a^2b + (-4 - 3)ab^2 + 6b^3\). 3. The result is \(6a^3 - 7a^2b - 7ab^2 + 6b^3\).

Answer

\(6a^3 - 7a^2b - 7ab^2 + 6b^3\)

Multiply polynomials 59

Expand and simplify. \((4x - 3)(2x^2 + x - 5)\)

Hints

- Use the distributive property to multiply every term in one factor by every term in the other. - Apply the sign rules carefully. - Combine terms only when they have the same power of \(x\). - Organizing the partial products in rows may help.

Solution

1. Form the partial products: \(8x^3 + 4x^2 - 20x - 6x^2 - 3x + 15\). 2. Combine terms with the same power of \(x\): \(8x^3 - 2x^2 - 23x + 15\).

Answer

\(8x^3 - 2x^2 - 23x + 15\)

Multiply polynomials 60

Find the product and simplify. \((2a^3 - 3a^2b + ab^2 - b^3)(a + 2b)\)

Hints

- Add exponents when multiplying powers with the same base. - Multiply every term in the first polynomial by both terms in the second. - Do not combine terms such as \(a^3b\) and \(a^2b^2\). - Verify that you formed all eight partial products.

Solution

1. Form all eight partial products: \(2a^4 + 4a^3b - 3a^3b - 6a^2b^2 + a^2b^2 + 2ab^3 - ab^3 - 2b^4\). 2. Combine like terms: \(2a^4 + a^3b - 5a^2b^2 + ab^3 - 2b^4\).

Answer

\(2a^4 + a^3b - 5a^2b^2 + ab^3 - 2b^4\)

Multiply polynomials 61

Expand and simplify. \((2y^3 - y^2 + 4y - 1)(y^2 - 3)\)

Hints

- A four-term polynomial times a two-term polynomial creates eight partial products. - Pay close attention when multiplying by \(-3\). - Add exponents when multiplying powers of \(y\). - Sort the partial products by exponent before combining.

Solution

1. Form the eight partial products: \(2y^5 - 6y^3 - y^4 + 3y^2 + 4y^3 - 12y - y^2 + 3\). 2. Arrange by descending powers and combine like terms: \(2y^5 - y^4 - 2y^3 + 2y^2 - 12y + 3\).

Answer

\(2y^5 - y^4 - 2y^3 + 2y^2 - 12y + 3\)

Multiply polynomials 62

Expand and simplify. \((x^2 - 2x + 1)(x^3 + 3x^2 - 5x + 2)\)

Hints

- Multiply each term in the first polynomial by the entire second polynomial. - Arrange partial products by powers of \(x\) before combining. - Remember that \(x^2 \cdot x^3 = x^5\). - Track the signs carefully, especially in the products involving \(-2x\).

Solution

1. Multiply each term in the first factor by the second polynomial: \(x^2(x^3 + 3x^2 - 5x + 2) = x^5 + 3x^4 - 5x^3 + 2x^2\), \(-2x(x^3 + 3x^2 - 5x + 2) = -2x^4 - 6x^3 + 10x^2 - 4x\), and \(x^3 + 3x^2 - 5x + 2\). 2. Combine equal powers: \(x^5 + x^4 - 10x^3 + 15x^2 - 9x + 2\).

Answer

\(x^5 + x^4 - 10x^3 + 15x^2 - 9x + 2\)

Multiply polynomials 63

Expand and simplify. \(\left(\frac{2}{3}a - \frac{1}{2}b\right)\left(\frac{3}{4}a + \frac{1}{3}b - 2\right)\)

Hints

- Multiply every term in the first factor by every term in the second. - Track the signs carefully. - Use a common denominator to combine fractional coefficients. - Combine only terms with matching variable factors.

Solution

1. Form the six partial products: \(\frac{1}{2}a^2 + \frac{2}{9}ab - \frac{4}{3}a - \frac{3}{8}ab - \frac{1}{6}b^2 + b\). 2. Combine the \(ab\)-terms: \(\frac{2}{9}ab - \frac{3}{8}ab = \left(\frac{16}{72} - \frac{27}{72}\right)ab = -\frac{11}{72}ab\). 3. The result is \(\frac{1}{2}a^2 - \frac{11}{72}ab - \frac{4}{3}a - \frac{1}{6}b^2 + b\).

Answer

\(\frac{1}{2}a^2 - \frac{11}{72}ab - \frac{4}{3}a - \frac{1}{6}b^2 + b\)

Multiply polynomials 64

Expand and simplify. \(\left(\frac{1}{2}x + \frac{1}{3}y - \frac{1}{4}z\right)\left(\frac{1}{2}x - \frac{1}{3}y + \frac{1}{4}z\right)\)

Hints

- Multiply every term in the first trinomial by every term in the second. - Apply the sign rules carefully. - Look for pairs of opposite terms after expanding. - Add fractions with the same denominator directly.

Solution

1. Form the nine partial products: \(\frac{1}{4}x^2 - \frac{1}{6}xy + \frac{1}{8}xz + \frac{1}{6}xy - \frac{1}{9}y^2 + \frac{1}{12}yz - \frac{1}{8}xz + \frac{1}{12}yz - \frac{1}{16}z^2\). 2. The \(xy\)-terms and \(xz\)-terms cancel. The \(yz\)-terms combine to \(\frac{1}{6}yz\). 3. The result is \(\frac{1}{4}x^2 - \frac{1}{9}y^2 + \frac{1}{6}yz - \frac{1}{16}z^2\).

Answer

\(\frac{1}{4}x^2 - \frac{1}{9}y^2 + \frac{1}{6}yz - \frac{1}{16}z^2\)

Multiply polynomials 65

Simplify the expression, then evaluate it for \(x = 0.75\). \((2x - 3)(x + 4) - (x - 2)(2x + 1)\)

Hints

- Simplifying first is more efficient than substituting immediately. - A minus sign before parentheses changes every sign inside. - Expand each product systematically. - Combine terms with the same power of \(x\).

Solution

1. Expand both products: \((2x^2 + 5x - 12) - (2x^2 - 3x - 2)\). 2. Remove the second parentheses and combine like terms: \(2x^2 + 5x - 12 - 2x^2 + 3x + 2 = 8x - 10\). 3. Substitute \(x = 0.75\): \(8 \cdot 0.75 - 10 = 6 - 10 = -4\).

Answer

Simplified expression: \(8x - 10\) Value at \(x = 0.75\): \(-4\)

Multiply polynomials 66

Let \(T(n) = (3n - 2)(n + 3) - (n - 1)(3n + 2) + (n + 2)(n - 2) - n^2\). Simplify the expression first, then evaluate it for \(n = -1\frac{1}{8}\).

Hints

- Break the expression into separate products and simplify each one. - One product is a difference of squares. - Pay close attention to the minus sign before the second product. - Convert the mixed number to an improper fraction before evaluating.

Solution

1. Expand the products: \((3n - 2)(n + 3) = 3n^2 + 7n - 6\), \((n - 1)(3n + 2) = 3n^2 - n - 2\), and \((n + 2)(n - 2) = n^2 - 4\). 2. Substitute these expressions and combine like terms: \((3n^2 + 7n - 6) - (3n^2 - n - 2) + (n^2 - 4) - n^2 = 8n - 8\). 3. Convert \(-1\frac{1}{8}\) to \(-\frac{9}{8}\), then evaluate: \(8 \cdot \left(-\frac{9}{8}\right) - 8 = -17\).

Answer

Simplified expression: \(8n - 8\) Value at \(n = -1\frac{1}{8}\): \(-17\)

Multiply polynomials 67

Before multiplying, arrange the terms in each factor in descending powers of \(x\). Then expand and simplify. \((5 + 3x^2 - 4x)(2x - 3 + x^2)\)

Hints

- First arrange each polynomial in descending powers. - Track the signs of all partial products. - Multiply every term in one factor by every term in the other. - Combine terms with the same power of \(x\).

Solution

1. Rewrite the factors as \((3x^2 - 4x + 5)(x^2 + 2x - 3)\). 2. Form the partial products: \(3x^4 + 6x^3 - 9x^2 - 4x^3 - 8x^2 + 12x + 5x^2 + 10x - 15\). 3. Combine like powers: \(3x^4 + 2x^3 - 12x^2 + 22x - 15\).

Answer

\(3x^4 + 2x^3 - 12x^2 + 22x - 15\)

Multiply polynomials 68

Before multiplying, arrange the first factor in descending powers of \(a\). Then expand and simplify. \((2b^2 - 3ab + a^2)(4a - 5b)\)

Hints

- Arrange the first polynomial by powers of \(a\). - Terms such as \(a^2b\) and \(ab^2\) are not like terms. - Track negative signs carefully. - Combine only terms with exactly matching variable factors.

Solution

1. Rewrite the first factor as \(a^2 - 3ab + 2b^2\). 2. Form the partial products: \(4a^3 - 5a^2b - 12a^2b + 15ab^2 + 8ab^2 - 10b^3\). 3. Combine like terms: \(4a^3 - 17a^2b + 23ab^2 - 10b^3\).

Answer

\(4a^3 - 17a^2b + 23ab^2 - 10b^3\)

Multiply polynomials 69

Expand and simplify. 1) \((2a^2 - 3ab + b^2)(4a - 5b)\) 2) \((x^2 - 3x + 2)(2x^2 + x - 4)\)

Hints

- Multiply every term in one factor by every term in the other. - Apply the sign rules carefully. - Arrange the result by powers before combining. - Add exponents when multiplying powers with the same base.

Solution

1. Form the partial products and combine like terms: \(8a^3 - 10a^2b - 12a^2b + 15ab^2 + 4ab^2 - 5b^3 = 8a^3 - 22a^2b + 19ab^2 - 5b^3\). 2. Expand and combine equal powers: \(2x^4 + x^3 - 4x^2 - 6x^3 - 3x^2 + 12x + 4x^2 + 2x - 8 = 2x^4 - 5x^3 - 3x^2 + 14x - 8\).

Answer

1) \(8a^3 - 22a^2b + 19ab^2 - 5b^3\) 2) \(2x^4 - 5x^3 - 3x^2 + 14x - 8\)

Multiply polynomials 70

Let \(T = (3x^2 - 2x + 1)(2x - 5)\). a) Expand and simplify \(T\). b) What is the coefficient of the \(x^2\)-term in the result? c) Evaluate \(T\) for \(x = 2\).

Hints

- A coefficient is the numerical factor of a variable term. - For part c), you may substitute into either the original product or the expanded expression. - Include the sign when combining the \(x^2\)-terms.

Solution

1. Expand: \(6x^3 - 15x^2 - 4x^2 + 10x + 2x - 5\). 2. Combine like terms: \(T = 6x^3 - 19x^2 + 12x - 5\). 3. The coefficient of \(x^2\) is \(-19\). 4. For \(x = 2\), \(T = 6 \cdot 2^3 - 19 \cdot 2^2 + 12 \cdot 2 - 5 = -9\).

Answer

a) \(6x^3 - 19x^2 + 12x - 5\) b) \(-19\) c) \(-9\)

Multiply polynomials 71

Let \(u = m + 2n\) and \(v = m - n\). Rewrite each expression using only \(m\) and \(n\), then simplify. 1) \(3u + 4v\) 2) \(2u - v\) 3) \(um - vm\) 4) \(uv\)

Hints

- A minus sign before parentheses changes every sign inside. - In part 3, factoring out \(m\) may simplify the setup. - When multiplying two binomials, multiply every term in one by every term in the other.

Solution

1. \(3(m + 2n) + 4(m - n) = 7m + 2n\). 2. \(2(m + 2n) - (m - n) = m + 5n\). 3. \(m(m + 2n) - m(m - n) = m^2 + 2mn - m^2 + mn = 3mn\). 4. \((m + 2n)(m - n) = m^2 - mn + 2mn - 2n^2 = m^2 + mn - 2n^2\).

Answer

1) \(7m + 2n\) 2) \(m + 5n\) 3) \(3mn\) 4) \(m^2 + mn - 2n^2\)

Multiply polynomials 72

A rectangle is \(7\,\text{in.}\) longer than it is wide. If the length is decreased by \(2\,\text{in.}\) and the width is increased by \(5\,\text{in.}\), the new figure is a square. The square's area is \(55\,\text{in.}^2\) greater than the rectangle's original area. Let \(x\) be the original width. a) Write and expand the original rectangle's area and the new square's area. b) Use the area difference to find the square's side length.

Hints

- Express the original length using the width. - Check why the two changed dimensions are equal. - Expand both areas and subtract before using the given difference.

Solution

1. The original rectangle has dimensions \(x\) and \(x+7\), so its area is \(x(x+7)=x^2+7x\). 2. After the changes, both dimensions are \(x+5\), so the square's area is \((x+5)^2=x^2+10x+25\). 3. The difference is \((x^2+10x+25)-(x^2+7x)=3x+25\). 4. Set \(3x+25=55\). Then \(x=10\). 5. The square's side length is \(x+5=15\,\text{in.}\).

Answer

a) Original area: \(x^2+7x\) Square area: \(x^2+10x+25\) b) \(3x+25=55\), so the square's side length is \(15\,\text{in.}\).

Multiply polynomials 73

A rectangle is \(8\,\text{in.}\) longer than it is wide. If the length is shortened by \(3\,\text{in.}\) and the width is increased by \(2\,\text{in.}\), the new rectangle has the same area as the original rectangle. Let \(b\) be the original width. a) Write and expand the original and new area expressions. b) Set the area expressions equal and find the original rectangle's perimeter.

Hints

- Express both dimensions using the original width. - Apply the changes before multiplying the new dimensions. - Use the equal-area condition only after both polynomials are written.

Solution

1. The original dimensions are \(b\) and \(b+8\), so the original area is \(b(b+8)=b^2+8b\). 2. The new dimensions are \(b+2\) and \(b+5\), so the new area is \((b+2)(b+5)=b^2+7b+10\). 3. Equal areas give \(b^2+8b=b^2+7b+10\). 4. Simplify: \(b=10\). 5. The original dimensions are \(10\,\text{in.}\) and \(18\,\text{in.}\), so the perimeter is \(2(10+18)=56\,\text{in.}\).

Answer

a) Original area: \(b^2+8b\) New area: \(b^2+7b+10\) b) \(b^2+8b=b^2+7b+10\), so the original perimeter is \(56\,\text{in.}\).

Multiply polynomials 74

Let \(x\) be the first of four consecutive odd integers \(x\), \(x+2\), \(x+4\), and \(x+6\). a) Expand and simplify \(D(x)=(x+4)(x+6)-x(x+2)\) to show how much greater the product of the last two integers is than the product of the first two. b) The product of the last two integers is \(272\) greater than the product of the first two. Use your simplified expression from part a) to find the four integers.

Hints

- Expand both products before comparing them. - Distribute the subtraction across the entire second product. - Notice what happens to the \(x^2\) terms. - Use the simplified expression, not trial-and-error testing, in part b.

Solution

1. Expand the first product: \((x+4)(x+6)=x^2+10x+24\). 2. Expand the second product: \(x(x+2)=x^2+2x\). 3. Subtract: \(D(x)=x^2+10x+24-(x^2+2x)=8x+24\). 4. For part b, set \(8x+24=272\). 5. Then \(8x=248\), so \(x=31\). 6. The four consecutive odd integers are \(31\), \(33\), \(35\), and \(37\).

Answer

a) \(D(x)=8x+24\) b) \(31, 33, 35, 37\)

Multiply polynomials 75

A rectangle is three times as long as it is wide. If the width is increased by \(2\,\text{in.}\) and the length is decreased by \(3\,\text{in.}\), the area increases by \(15\,\text{in.}^2\). Let \(x\) be the original width. a) Write and expand the original and new area expressions. b) Subtract the original area from the new area and use the given increase to find the original dimensions.

Hints

- Express the length in terms of the width. - Apply both dimension changes before expanding. - Compare the two expanded area expressions.

Solution

1. The original dimensions are \(x\) and \(3x\), so the original area is \(3x^2\). 2. The new dimensions are \(x+2\) and \(3x-3\), so the new area is \((x+2)(3x-3)=3x^2+3x-6\). 3. The increase is \(3x-6\). 4. Set \(3x-6=15\), giving \(x=7\). 5. The original dimensions were \(7\,\text{in.}\) by \(21\,\text{in.}\).

Answer

a) Original area: \(3x^2\) New area: \(3x^2+3x-6\) b) Area increase: \(3x-6\) \(3x-6=15\), so the original dimensions were \(7\,\text{in.}\) by \(21\,\text{in.}\).

Multiply polynomials 76

Expand and simplify. \(4a(2a - 3b) - (a - b)(3a + 2b) - 5a^2\)

Hints

- Pay close attention to the minus sign before the binomial product. - Expand the product completely before applying the outside subtraction. - Group terms with the same variable factors.

Solution

1. Expand the first product: \(4a(2a - 3b) = 8a^2 - 12ab\). 2. Expand the binomial product: \((a - b)(3a + 2b) = 3a^2 - ab - 2b^2\). 3. Subtract this product and combine like terms: \(8a^2 - 12ab - 3a^2 + ab + 2b^2 - 5a^2 = 2b^2 - 11ab\).

Answer

\(2b^2 - 11ab\)

Multiply polynomials 77

A rectangular garden bed is \(5\,\text{ft}\) longer than it is wide. If its width is increased by \(2\,\text{ft}\) and its length is increased by \(3\,\text{ft}\), its area increases by \(41\,\text{ft}^2\). Let \(x\) be the original width. a) Write and expand the original and new area expressions. b) Find the polynomial expression for the area increase, then use it to find the original dimensions.

Hints

- Express the original length using the width. - Expand the new area after both dimensions change. - Subtract corresponding polynomial terms before solving.

Solution

1. The original dimensions are \(x\) and \(x+5\), so the original area is \(x^2+5x\). 2. The new dimensions are \(x+2\) and \(x+8\), so the new area is \((x+2)(x+8)=x^2+10x+16\). 3. The increase is \(5x+16\). 4. Set \(5x+16=41\), giving \(x=5\). 5. The original dimensions were \(5\,\text{ft}\) by \(10\,\text{ft}\).

Answer

a) Original area: \(x^2+5x\) New area: \(x^2+10x+16\) b) Area increase: \(5x+16\) \(5x+16=41\), so the original dimensions were \(5\,\text{ft}\) by \(10\,\text{ft}\).

Multiply polynomials 78

A square playground is redesigned as a rectangle by increasing one side by \(6\,\text{ft}\) and decreasing the other side by \(4\,\text{ft}\). The new area is \(16\,\text{ft}^2\) greater than the original area. a) Let \(s\) be the original side length. Write and expand the original and redesigned area expressions, subtract them, and find \(s\). b) Someone claims, “If one side is increased by the same amount that the other side is decreased, the area always stays the same.” Use polynomial expressions with changes of \(5\,\text{ft}\) to test the claim for the side length from part a).

Hints

- Expand the redesigned rectangle before comparing areas. - The quadratic terms cancel in the area difference. - For equal opposite changes, compare \((s+d)(s-d)\) with \(s^2\).

Solution

1. The original area is \(s^2\). 2. The redesigned area is \((s+6)(s-4)=s^2+2s-24\). 3. The increase is \(2s-24\). Set \(2s-24=16\), so \(s=20\). 4. For equal changes of \(5\,\text{ft}\), the new area is \((s+5)(s-5)=s^2-25\). 5. At \(s=20\), the new area is \(400-25=375\,\text{ft}^2\), not \(400\,\text{ft}^2\). The claim is false.

Answer

a) Original area: \(s^2\) Redesigned area: \(s^2+2s-24\) Area increase: \(2s-24\) \(s=20\,\text{ft}\) b) \((s+5)(s-5)=s^2-25\). For \(s=20\), the area is \(375\,\text{ft}^2\), so the claim is false.

Multiply polynomials 79

Let \(B(a) = 2a(3a - 1) - (a + 2)(6a - 5)\). Simplify the expression completely, then evaluate it for \(a = -\frac{2}{3}\).

Hints

- Expand the two products separately. - Keep the second product in parentheses when subtracting it. - Combine the \(a^2\)-terms, \(a\)-terms, and constants separately. - Substitute only after simplifying completely.

Solution

1. Expand the first product: \(2a(3a - 1) = 6a^2 - 2a\). 2. Expand the second product: \((a + 2)(6a - 5) = 6a^2 + 7a - 10\). 3. Subtract and combine like terms: \(6a^2 - 2a - (6a^2 + 7a - 10) = -9a + 10\). 4. Evaluate: \(-9 \cdot \left(-\frac{2}{3}\right) + 10 = 16\).

Answer

\(16\)

Multiply polynomials 80

For numbers close to \(1\), the approximation \((1 + a)(1 + b) \approx 1 + a + b\) can be useful when \(a\) and \(b\) are very small. a) Expand \((1 + a)(1 + b)\). Which term is omitted by the approximation? b) Use the approximation to estimate \(1.002 \cdot 1.006\). c) Find the exact product and the difference between the exact value and the estimate.

Hints

- Multiply every term in the first binomial by every term in the second. - Write each decimal as \(1\) plus a small amount. - The error is the difference between the exact product and the estimate.

Solution

1. Expand: \((1 + a)(1 + b) = 1 + a + b + ab\). The approximation omits \(ab\). 2. Here \(a = 0.002\) and \(b = 0.006\), so the estimate is \(1 + 0.002 + 0.006 = 1.008\). 3. The exact product is \(1.002 \cdot 1.006 = 1.008012\). 4. The difference is \(1.008012 - 1.008 = 0.000012\).

Answer

a) \((1 + a)(1 + b) = 1 + a + b + ab\); the omitted term is \(ab\) b) \(1.008\) c) Exact product: \(1.008012\); difference: \(0.000012\)

Multiply polynomials 81

Simplify the expression. Then evaluate it for the given values: \(T(a, b) = 4a(a - 2b) + 2b(4a + b) - 3a^2\) Use \(a = 1.2\) and \(b = -0.5\).

Hints

- Distribute each factor across its parentheses. - Watch the signs as you combine like terms. - Look for terms that cancel. - Substitute the values after simplifying.

Solution

1. Distribute: \(4a(a - 2b) = 4a^2 - 8ab\) and \(2b(4a + b) = 8ab + 2b^2\). 2. Combine the expressions: \(4a^2 - 8ab + 8ab + 2b^2 - 3a^2\). 3. The opposite \(ab\)-terms cancel, so \(T(a,b) = a^2 + 2b^2\). 4. Substitute: \((1.2)^2 + 2 \cdot (-0.5)^2 = 1.44 + 2 \cdot 0.25 = 1.94\).

Answer

Simplified expression: \(a^2 + 2b^2\) Value: \(1.94\)

Multiply polynomials 82

Consider the two expressions \(A = x(2x + y) - y(x - 3y)\) \(B = 2(x^2 + y^2) + y^2\). a) Use algebraic properties to show that \(A\) and \(B\) are equivalent. b) Find their common value when \(x = -5\) and \(y = \frac{1}{3}\).

Hints

- Simplify each expression separately and compare the results. - Remember that \(xy = yx\). - A minus sign before parentheses changes the sign of each term when the parentheses are removed. - Substitute into the simpler equivalent expression.

Solution

1. Simplify \(A\): \(A = 2x^2 + xy - xy + 3y^2 = 2x^2 + 3y^2\). 2. Simplify \(B\): \(B = 2x^2 + 2y^2 + y^2 = 2x^2 + 3y^2\). 3. Because both expressions simplify to \(2x^2 + 3y^2\), they are equivalent. 4. Substitute: \(2 \cdot (-5)^2 + 3 \cdot \left(\frac{1}{3}\right)^2 = 50 + \frac{1}{3} = \frac{151}{3}\).

Answer

a) Both expressions simplify to \(2x^2 + 3y^2\). b) \(\frac{151}{3}\), or \(50\frac{1}{3}\)

Multiply polynomials 83

An open-top box is made from a square sheet of cardboard by cutting a \(4\,\text{in.}\)-by-\(4\,\text{in.}\) square from each corner and folding up the sides. The area of the box's base is \(160\,\text{in.}^2\) less than the area of the original sheet. Let \(a\) be the original side length. a) Write the original sheet area and the box-base area as polynomial expressions. Expand the base area and simplify their difference. b) Use the area difference to find \(a\), then find the box's volume.

Hints

- Cutting a square from both ends reduces each base dimension by twice the cutout side. - Expand the square before subtracting areas. - Use the solved side length to determine the base dimensions and height.

Solution

1. The original area is \(a^2\). 2. The base side length is \(a-8\), so the base area is \((a-8)^2=a^2-16a+64\). 3. The area difference is \(a^2-(a^2-16a+64)=16a-64\). 4. Set \(16a-64=160\), so \(a=14\). 5. The base is \(6\,\text{in.}\) by \(6\,\text{in.}\), and the height is \(4\,\text{in.}\). 6. The volume is \(6\cdot6\cdot4=144\,\text{in.}^3\).

Answer

a) Original area: \(a^2\) Base area: \((a-8)^2=a^2-16a+64\) Area difference: \(16a-64\) b) \(a=14\,\text{in.}\); volume \(=144\,\text{in.}^3\)

Multiply polynomials 84

A rectangular garden bed is \(8\,\text{m}\) longer than it is wide. The length is decreased by \(3\,\text{m}\), and the width is increased by \(3\,\text{m}\). Let \(b\) be the original width. Write and expand both area expressions, then subtract them to find the area change.

Hints

- Express the starting length using the width \(b\). - Apply both changes to the dimensions before finding the new area. - Expand the products and subtract the original area from the new area.

Solution

1. The original length is \(b+8\), so the original area is \(A_1=b(b+8)=b^2+8b\). 2. The new width and length are \(b+3\) and \(b+5\); the new area is \(A_2=(b+3)(b+5)=b^2+8b+15\). 3. Subtracting gives \(A_2-A_1=15\), so the area increases by \(15\,\text{m}^2\).

Answer

Original area: \(b(b+8)=b^2+8b\) New area: \((b+3)(b+5)=b^2+8b+15\) Difference: \(A_2-A_1=15\,\text{m}^2\), an increase

Multiply polynomials 85

A rectangular parking lot is \(12\,\text{m}\) longer than it is wide. A second parking lot is square, and its side length is the mean of the rectangular lot's length and width. a) Let \(b\) represent the rectangular lot's width. Write an expression for the area of each parking lot. b) Which parking lot has the greater area, and what is the exact difference?

Hints

- The mean of two values is their sum divided by \(2\). - Express the square's side length in terms of \(b\). - Expand both area expressions before comparing. - Identify the term that differs between the two results.

Solution

1. The rectangular lot has width \(b\) and length \(b+12\), so \(A_R=b(b+12)=b^2+12b\). 2. The square's side length is the mean: \(s=\frac{b+(b+12)}{2}=b+6\). 3. Its area is \(A_S=(b+6)^2=b^2+12b+36\). 4. Since \(A_S-A_R=36\), the square lot has \(36\,\text{m}^2\) more area.

Answer

a) Rectangle: \(b(b+12)=b^2+12b\); square: \((b+6)^2=b^2+12b+36\) b) The square parking lot has \(36\,\text{m}^2\) more area.

Multiply polynomials 86

Expand and simplify. Assume \(n\), \(k\), and \(m\) are nonnegative integers. 1) \(4a^n(a^3 - 2a^n)\) 2) \(-3x^k(2x^{k+1} + 4x)\) 3) \(\frac{1}{3}y^m(9y^{2m} - 6y)\)

Hints

- Multiply the outside factor by every term inside the parentheses. - Add exponents when multiplying powers with the same base. - Track the negative sign during distribution.

Solution

1. Distribute and add exponents: \(4a^{n+3} - 8a^{2n}\). 2. Distribute: \(-6x^{k+(k+1)} - 12x^{k+1} = -6x^{2k+1} - 12x^{k+1}\). 3. Distribute: \(3y^{m+2m} - 2y^{m+1} = 3y^{3m} - 2y^{m+1}\).

Answer

1) \(4a^{n+3} - 8a^{2n}\) 2) \(-6x^{2k+1} - 12x^{k+1}\) 3) \(3y^{3m} - 2y^{m+1}\)

Multiply polynomials 87

A square has side length \(s\) inches. One side is increased by \(2\,\text{in.}\), and the other is increased by \(5\,\text{in.}\), forming a rectangle. The rectangle's area is \(45\,\text{in.}^2\) greater than the square's area. a) Write and expand the square and rectangle area expressions. b) Subtract the square area from the rectangle area and use the given difference to find \(s\).

Hints

- Write the two new side lengths in terms of \(s\). - Expand the rectangle area before comparing it with the square. - The \(s^2\) terms cancel in the difference.

Solution

1. The square area is \(s^2\). 2. The rectangle area is \((s+2)(s+5)=s^2+7s+10\). 3. The area difference is \(7s+10\). 4. Set \(7s+10=45\), so \(s=5\).

Answer

a) Square area: \(s^2\) Rectangle area: \(s^2+7s+10\) b) Area difference: \(7s+10\) \(7s+10=45\), so \(s=5\,\text{in.}\).

Multiply polynomials 88

A rectangular garden is \(5\,\text{ft}\) longer than it is wide. If its width increases by \(2\,\text{ft}\) while its length stays the same, its area increases by \(30\,\text{ft}^2\). Let \(b\) be the original width. Write and expand both area products, subtract them to form an area-increase equation, then find the original dimensions.

Hints

- Write the original length in terms of \(b\). - The width changes, but the length does not. - Subtract the original expanded area from the new one. - Use the given increase only after finding the difference.

Solution

1. The original length is \(b+5\), so the original area is \(b(b+5)=b^2+5b\). 2. The new area is \((b+2)(b+5)=b^2+7b+10\). 3. The increase is \((b^2+7b+10)-(b^2+5b)=2b+10\). 4. Set \(2b+10=30\); then \(b=10\,\text{ft}\). The length was \(b+5=15\,\text{ft}\).

Answer

Original area: \(b(b+5)=b^2+5b\) New area: \((b+2)(b+5)=b^2+7b+10\) Increase equation: \(2b+10=30\) Original dimensions: \(10\,\text{ft}\) wide and \(15\,\text{ft}\) long

Multiply polynomials 89

Verify the identity by expanding and combining like terms. \((x + 3)(x^2 - 3x + 9) = x^3 + 27\)

Hints

- Multiply each term in the first factor by every term in the second factor. - Track the signs carefully. - Look for opposite terms that cancel.

Solution

1. Distribute: \((x + 3)(x^2 - 3x + 9) = x(x^2 - 3x + 9) + 3(x^2 - 3x + 9)\). 2. Multiply: \(x^3 - 3x^2 + 9x + 3x^2 - 9x + 27\). 3. The quadratic terms and linear terms cancel, leaving \(x^3 + 27\). Therefore, the identity is true.

Answer

The left side simplifies to \(x^3 + 27\), so the identity is verified.

Multiply polynomials 90

The product \((x+k)(2x-3)\) has an \(x\)-term of \(5x\). Find \(k\), then find the constant term of the product. Show the multiplication that determines each answer.

Hints

- Expand the factors while leaving \(k\) unknown. - Which two partial products contribute to the \(x\)-term? - Once you determine \(k\), check the product of the constant terms.

Solution

1. Expanding gives \((x+k)(2x-3)=2x^2+(2k-3)x-3k\). 2. The coefficient of \(x\) is \(5\), so \(2k-3=5\), giving \(k=4\). 3. The constant term is \(-3k=-12\).

Answer

The \(x\)-term is \((2k-3)x=5x\), so \(2k-3=5\) and \(k=4\). The constant term is \(k\cdot(-3)=-3k=-12\).

Multiply polynomials 91

Examine the accuracy of \((1 + a)(1 + b) \approx 1 + a + b\) in these cases: Case I: \(1.003 \cdot 1.003\) Case II: \(1.003 \cdot 0.997\) a) Find both approximate values. b) Find both exact products. c) Compare the absolute errors. Explain your observation using the omitted term \(ab\).

Hints

- In Case II, the second number is less than \(1\), so \(b\) is negative. - Compare the exact expansion with the approximation to identify the error term. - Compute the absolute difference between each estimate and exact product.

Solution

1. Case I has \(a = 0.003\), \(b = 0.003\), so the estimate is \(1.006\). 2. Case II has \(a = 0.003\), \(b = -0.003\), so the estimate is \(1.000\). 3. The exact products are \(1.003 \cdot 1.003 = 1.006009\) and \(1.003 \cdot 0.997 = 0.999991\). 4. Each absolute error is \(0.000009\). The omitted products are \(0.003 \cdot 0.003 = 0.000009\) and \(0.003 \cdot (-0.003) = -0.000009\), which have the same absolute value.

Answer

a) Case I: \(1.006\); Case II: \(1.000\) b) Case I: \(1.006009\); Case II: \(0.999991\) c) Both absolute errors are \(0.000009\) because the omitted terms \(ab\) have equal absolute values.

Multiply polynomials 92

A rectangular sheet of cardboard is exactly twice as long as it is wide. An open-top box is made by cutting a \(2\,\text{in.}\)-by-\(2\,\text{in.}\) square from each corner and folding up the sides. The original sheet has \(44\,\text{in.}^2\) more area than the box's base. Let \(b\) be the original width. a) Write and expand the original sheet area and the box-base area. Simplify their difference. b) Use the area difference to find the original dimensions and the volume of the box.

Hints

- Express the original length as twice the width. - Each base dimension loses \(4\,\text{in.}\) in total. - Expand the base-area product and subtract it from the original area.

Solution

1. The original sheet is \(b\) by \(2b\), so its area is \(2b^2\). 2. The base dimensions are \(b-4\) and \(2b-4\), so the base area is \((b-4)(2b-4)=2b^2-12b+16\). 3. The area difference is \(2b^2-(2b^2-12b+16)=12b-16\). 4. Set \(12b-16=44\), so \(b=5\). 5. The original sheet is \(5\,\text{in.}\) by \(10\,\text{in.}\). 6. The base is \(1\,\text{in.}\) by \(6\,\text{in.}\), with height \(2\,\text{in.}\), so the volume is \(12\,\text{in.}^3\).

Answer

a) Original area: \(2b^2\) Base area: \(2b^2-12b+16\) Area difference: \(12b-16\) b) Original dimensions: \(5\,\text{in.}\) by \(10\,\text{in.}\) Volume: \(12\,\text{in.}^3\)

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