Free Math Worksheets

- Use the pattern for the square of a difference. - Square both the coefficient and variable in \(2p\). - Include the middle product term.

1. Use \((a - b)^2 = a^2 - 2ab + b^2\). 2. Substitute \(a = 2p\) and \(b = q\): \((2p)^2 - 2(2p)(q) + q^2\). 3. Simplify to \(4p^2 - 4pq + q^2\).

c) \(4p^2 - 4pq + q^2\)

- Identify the matching binomial identity. - Square both numerator and denominator of a fractional term. - Recognize conjugate binomials in part c).

1. a) \(\left(\frac{2}{5}x\right)^2 + 2\left(\frac{2}{5}x\right)(10) + 10^2 = \frac{4}{25}x^2 + 8x + 100\). 2. b) \((1.5a)^2 - 2(1.5a)(2b) + (2b)^2 = 2.25a^2 - 6ab + 4b^2\). 3. c) Use the difference of squares: \((8k)^2 - (3m)^2 = 64k^2 - 9m^2\).

a) \(\frac{4}{25}x^2 + 8x + 100\) b) \(2.25a^2 - 6ab + 4b^2\) c) \(64k^2 - 9m^2\)

- Match each expression to a binomial identity. - Square every factor in a monomial. - Identify the two quantities that play the roles of \(A\) and \(B\).

1. a) \((3a)^2 + 2(3a)(5b) + (5b)^2 = 9a^2 + 30ab + 25b^2\). 2. b) \((4x^2)^2 - 2(4x^2)(7) + 7^2 = 16x^4 - 56x^2 + 49\). 3. c) Use the difference of squares: \((2y^3)^2 - (3x^2)^2 = 4y^6 - 9x^4\).

a) \(9a^2 + 30ab + 25b^2\) b) \(16x^4 - 56x^2 + 49\) c) \(4y^6 - 9x^4\)

- Find the midpoint of each pair of factors. - Rewrite the factors as a sum and difference of the same two values. - Use \((a + b)(a - b) = a^2 - b^2\).

1. a) \((20 + 1)(20 - 1) = 20^2 - 1^2 = 399\). 2. b) \((40 + 2)(40 - 2) = 40^2 - 2^2 = 1596\). 3. c) \((100 + 3)(100 - 3) = 100^2 - 3^2 = 9991\).

a) \(399\) b) \(1596\) c) \(9991\)

- Factor \(a^2 - b^2\) as \((a - b)(a + b)\). - Pay attention to the sign of the first factor. - Multiplying the resulting factors may be easier than squaring first.

1. a) \((45 - 35)(45 + 35) = 10 \cdot 80 = 800\). 2. b) \((16 - 24)(16 + 24) = -8 \cdot 40 = -320\). 3. c) \((101 - 99)(101 + 99) = 2 \cdot 200 = 400\).

a) \(800\) b) \(-320\) c) \(400\)

- Isolate the squared expression. - Recall the possible values of a square of a real number. - Compare those values with the right side of the equation.

1. Subtract \(5\) from both sides: \(3(x - 2)^2 = -3\). 2. Divide by \(3\): \((x - 2)^2 = -1\). 3. The square of a real number is always nonnegative, so \((x - 2)^2\) cannot equal \(-1\). 4. Therefore, the equation has no real solution.

There is no real solution because the equation is equivalent to \((x - 2)^2 = -1\), and a real square cannot be negative.

- Identify the two terms in each binomial. - Square each term and include twice their product. - Use the sign between the terms to determine the sign of the middle term.

1. a) \(x^2 + 24x + 144\). 2. b) \(100 - 20y + y^2\). 3. c) \(16a^2 + 24ab + 9b^2\). 4. d) \(36m^2 - 60m + 25\).

a) \(x^2 + 24x + 144\) b) \(100 - 20y + y^2\) c) \(16a^2 + 24ab + 9b^2\) d) \(36m^2 - 60m + 25\)

- Identify the two terms in each binomial. - Square both terms and include twice their product. - Square the numerator and denominator of fractional terms.

1. \(a^2 + 2a\left(\frac{2}{3}\right) + \left(\frac{2}{3}\right)^2 = a^2 + \frac{4}{3}a + \frac{4}{9}\). 2. \((0.4x)^2 - 2(0.4x)(5) + 25 = 0.16x^2 - 4x + 25\). 3. \(\left(\frac{1}{2}y\right)^2 + 2\left(\frac{1}{2}y\right)\left(\frac{1}{4}z\right) + \left(\frac{1}{4}z\right)^2 = \frac{1}{4}y^2 + \frac{1}{4}yz + \frac{1}{16}z^2\).

1) \(a^2 + \frac{4}{3}a + \frac{4}{9}\) 2) \(0.16x^2 - 4x + 25\) 3) \(\frac{1}{4}y^2 + \frac{1}{4}yz + \frac{1}{16}z^2\)

- Choose a nearby multiple of \(10\), \(50\), or \(100\). - Write the original number as a sum or difference from that benchmark. - Keep track of place value when squaring a decimal.

1. a) Write \(21=20+1\): \((20+1)^2=20^2+2 \cdot 20 \cdot 1+1^2=400+40+1=441\). 2. b) Write \(49=50-1\): \((50-1)^2=50^2-2 \cdot 50 \cdot 1+1^2=2500-100+1=2401\). 3. c) Write \(10.1=10+0.1\): \((10+0.1)^2=10^2+2 \cdot 10 \cdot 0.1+(0.1)^2=100+2+0.01=102.01\). 4. d) Write \(199=200-1\): \((200-1)^2=200^2-2 \cdot 200 \cdot 1+1^2=40{,}000-400+1=39{,}601\).

a) \(441\) b) \(2401\) c) \(102.01\) d) \(39{,}601\)

- Distribute the factor outside each set of parentheses. - Pay close attention to the negative sign in the second identity. - Compare the resulting trinomial with a binomial-square pattern.

1. Distribute on the left: \(a^2+b(2a+b)=a^2+2ab+b^2=(a+b)^2\). 2. Distribute the negative product: \(a^2-b(2a-b)=a^2-2ab+b^2=(a-b)^2\). 3. In each case, the expanded left side matches the corresponding binomial-square identity.

1) \(a^2+b(2a+b)=a^2+2ab+b^2=(a+b)^2\) 2) \(a^2-b(2a-b)=a^2-2ab+b^2=(a-b)^2\)

- Expand each binomial square. - Distribute the subtraction sign across every term in the second trinomial. - Identify which terms cancel.

1. Expand both squares: \((a+b)^2=a^2+2ab+b^2\) and \((a-b)^2=a^2-2ab+b^2\). 2. Subtract the second trinomial: \(a^2+2ab+b^2-a^2+2ab-b^2\). 3. Combine like terms to obtain \(4ab\). Therefore, the identity is true for all real \(a\) and \(b\).

The identity is true because \((a+b)^2-(a-b)^2=4ab\).

- Rewrite each difference of squares as a product. - Subtract once and add once. - Look for factors that are easy to multiply mentally.

1. a) \(\left(5\frac{1}{2}-4\frac{1}{2}\right)\left(5\frac{1}{2}+4\frac{1}{2}\right)=1\cdot10=10\). 2. b) \((205-5)(205+5)=200\cdot210=42{,}000\). 3. c) \((8.7-1.3)(8.7+1.3)=7.4\cdot10=74\).

a) \(10\) b) \(42{,}000\) c) \(74\)

- Evaluate each expression separately using \(A^2-B^2=(A-B)(A+B)\). - Compare the two products rather than calculating the original squares.

1. \(A=(55-45)(55+45)=10\cdot100=1000\). 2. \(B=(102.5-97.5)(102.5+97.5)=5\cdot200=1000\). 3. Therefore, \(A=B\).

\(A=B=1000\)

- Find the square roots of the first and last terms. - Check whether the middle term is twice their product. - Use the middle sign to choose a sum or difference.

1. a) \(x^2+12x+36=x^2+2\cdot x\cdot6+6^2=(x+6)^2\). 2. b) \(25a^2-20ab+4b^2=(5a)^2-2\cdot5a\cdot2b+(2b)^2=(5a-2b)^2\). 3. c) \(w^2-18w+81=w^2-2\cdot w\cdot9+9^2=(w-9)^2\). 4. d) \(49z^2+14z+1=(7z)^2+2\cdot7z\cdot1+1^2=(7z+1)^2\).

a) \((x+6)^2\) b) \((5a-2b)^2\) c) \((w-9)^2\) d) \((7z+1)^2\)

- Take square roots of the first and last terms. - Verify that twice their product equals the middle term. - Use the sign of the middle term to choose the binomial sign.

1. a) The outer terms are \((0.7x)^2\) and \(y^2\), and \(2\cdot0.7x\cdot y=1.4xy\). Thus the expression is \((0.7x+y)^2\). 2. b) The outer terms are \((a^2)^2\) and \((0.3b)^2\), and \(-2\cdot a^2\cdot0.3b=-0.6a^2b\). Thus the expression is \((a^2-0.3b)^2\). 3. c) The outer terms are \((1.2u)^2\) and \(1^2\), and \(-2\cdot1.2u\cdot1=-2.4u\). Thus the expression is \((1.2u-1)^2\).

a) \((0.7x+y)^2\) b) \((a^2-0.3b)^2\) c) \((1.2u-1)^2\)

- Match each expression to a sum-square or difference-square identity. - Square every coefficient and variable factor. - Apply the power-of-a-power rule to existing exponents.

1. a) \((5x^3+2y^2)^2=25x^6+20x^3y^2+4y^4\). 2. b) The square terms are \(\frac{1}{9}a^4b^2\) and \(36a^2b^4\). The middle term is \(-2\cdot\left(\frac{1}{3}a^2b\right)\cdot6ab^2=-4a^3b^3\). Therefore, the result is \(\frac{1}{9}a^4b^2-4a^3b^3+36a^2b^4\).

a) \(25x^6+20x^3y^2+4y^4\) b) \(\frac{1}{9}a^4b^2-4a^3b^3+36a^2b^4\)

- Match each expression to a square of a sum, square of a difference, or difference of squares. - Square both the coefficient and variable factor in a monomial. - Include twice the product in a binomial square.

1. a) \((3x+4)^2=(3x)^2+2\cdot3x\cdot4+4^2=9x^2+24x+16\). 2. b) \((7-2b)^2=7^2-2\cdot7\cdot2b+(2b)^2=49-28b+4b^2\). 3. c) \((5k+1)(5k-1)=(5k)^2-1^2=25k^2-1\).

a) \(9x^2+24x+16\) b) \(49-28b+4b^2\) c) \(25k^2-1\)

- Identify square terms at the beginning and end. - For a trinomial, verify that the middle term is twice the product of the binomial terms. - For two square terms separated by subtraction, use the difference-of-squares identity.

1. a) \(m^2+12m+36=m^2+2\cdot m\cdot6+6^2=(m+6)^2\). 2. b) \(25n^2-40n+16=(5n)^2-2\cdot5n\cdot4+4^2=(5n-4)^2\). 3. c) \(49-9p^2=7^2-(3p)^2=(7-3p)(7+3p)\).

a) \((m+6)^2\) b) \((5n-4)^2\) c) \((7-3p)(7+3p)\)

- Evaluate the quantity inside parentheses before squaring. - Interpret the word “always” as a claim about every allowable input. - Distinguish an example from a counterexample.

1. a) The left side is \((3 + 4)^2 = 49\), while the right side is \(3^2 + 4^2 = 25\). Since \(49 \ne 25\), the formula is false. 2. b) A universal formula must work for every allowable pair of values. One successful example cannot prove a universal statement, while the counterexample in part a) is enough to disprove it.

a) The left side is \(49\), and the right side is \(25\), so the formula is false. b) One special case does not prove a statement for all values. The counterexample \(a = 3\), \(b = 4\) disproves the formula.

- Check the two square terms and the doubled product term. - Verify whether the factor \(2\) appears in the middle term. - Track the sign of the middle term.

1. The middle term is missing, and the constant term must be positive: \((x - 6)^2 = x^2 - 12x + 36\). 2. The middle term is \(2(2a)(5) = 20a\), so \((2a + 5)^2 = 4a^2 + 20a + 25\). 3. The expansion is correct because \(2(0.5y)(4) = 4y\).

1) Incorrect; \(x^2 - 12x + 36\) 2) Incorrect; \(4a^2 + 20a + 25\) 3) Correct

- Identify perfect-square first and last terms. - Check whether the middle term is twice the product of their square roots. - A difference of two squares factors into conjugates.

1. a) The first and last terms are \((3x)^2\) and \((2y)^2\), and the middle term is \(2(3x)(2y)\). Thus, \((3x + 2y)^2\). 2. b) The first and last terms are \((5a^2)^2\) and \((4b)^2\), and the middle term is \(-2(5a^2)(4b)\). Thus, \((5a^2 - 4b)^2\). 3. c) This is a difference of squares: \((7z + 9)(7z - 9)\).

a) \((3x + 2y)^2\) b) \((5a^2 - 4b)^2\) c) \((7z + 9)(7z - 9)\)

- Use length times width for each area. - Recognize a product of a sum and a difference. - Compare \(s^2\) with \(s^2 - 25\).

1. The square has area \(s^2\,\text{ft}^2\). 2. The rectangle has dimensions \(s + 5\) and \(s - 5\), so its area is \((s + 5)(s - 5)\,\text{ft}^2\). 3. Using the difference of squares, \((s + 5)(s - 5) = s^2 - 25\). 4. Therefore, the square has \(25\,\text{ft}^2\) more area.

a) Square: \(s^2\,\text{ft}^2\); rectangle: \((s + 5)(s - 5)\,\text{ft}^2\) b) The square has \(25\,\text{ft}^2\) more area.

- Substitute the given value into both sides. - Expand \((x + 5)^2\) completely. - Solve the resulting equation after canceling common terms.

1. a) At \(x = 1\), the left side is \(36\), and the right side is \(26\), so the claim is false in general. 2. b) Expand the left side: \(x^2 + 10x + 25 = x^2 + 25\). Thus, \(10x = 0\), so \(x = 0\). 3. c) The student omitted the middle term \(2(x)(5) = 10x\).

a) \(36 \ne 26\) b) \(x = 0\) c) The missing term is \(10x\).

- Look at the form of the original equation before expanding. - Expand both approaches only far enough to compare their standard forms. - Count which operations each method requires.

1. For part a, divide by \(2\): \((x - 3)^2 = 25\). Then \(x - 3 = \pm 5\), so \(x = 8\) or \(x = -2\). 2. For Anna's method, expand: \(2(x^2 - 6x + 9) = 50\). This gives \(2x^2 - 12x - 32 = 0\), and dividing by \(2\) gives \(x^2 - 6x - 16 = 0\). 3. Starting from Ben's first step, \((x - 3)^2 = 25\) becomes \(x^2 - 6x + 9 = 25\), so \(x^2 - 6x - 16 = 0\). Both approaches produce the same standard-form equation. 4. Ben's method uses the existing squared-binomial structure, avoiding expansion and the quadratic formula.

a) \(x \in \{-2, 8\}\) b) Both approaches lead to \(x^2 - 6x - 16 = 0\). c) Ben's method uses the squared-binomial form directly, so it avoids unnecessary expansion and formula work.

- Square coefficients and variable powers separately. - Multiply exponents when raising a power to a power. - Include the factor \(2\) in the middle term.

1. a) The square terms are \(\frac{9}{16}a^4\) and \(\frac{1}{9}b^2\), and the middle term is \(2\left(\frac{3}{4}a^2\right)\left(\frac{1}{3}b\right) = \frac{1}{2}a^2b\). 2. b) The square terms are \(2.25z^6\) and \(4w^4\), and the middle term is \(-2(1.5z^3)(2w^2) = -6z^3w^2\).

a) \(\frac{9}{16}a^4 + \frac{1}{2}a^2b + \frac{1}{9}b^2\) b) \(2.25z^6 - 6z^3w^2 + 4w^4\)

- Consider what happens when an entire negative expression is squared. - Identify the two terms in the binomial. - Square coefficients and variable powers separately. - Remember the factor \(2\) in the middle term.

1. Squaring a negative expression gives the same result as squaring its opposite: \((-1.5a^2b - 0.4ab^2)^2 = (1.5a^2b + 0.4ab^2)^2\). 2. Apply \((A+B)^2=A^2+2AB+B^2\). 3. The square terms are \((1.5a^2b)^2=2.25a^4b^2\) and \((0.4ab^2)^2=0.16a^2b^4\). 4. The middle term is \(2(1.5a^2b)(0.4ab^2)=1.2a^3b^3\). 5. Therefore, the simplified expression is \(2.25a^4b^2+1.2a^3b^3+0.16a^2b^4\).

\(2.25a^4b^2+1.2a^3b^3+0.16a^2b^4\)

- Match each expression to either \((A-B)^2\) or \((A+B)^2\). - Square each term and include twice their product. - Apply the power-of-a-power rule when squaring \(x^2\).

1. For \(\left(\frac{2}{3}x-6\right)^2\), the square terms are \(\frac{4}{9}x^2\) and \(36\), and the middle term is \(-2\left(\frac{2}{3}x\right)(6)=-8x\). Thus the result is \(\frac{4}{9}x^2-8x+36\). 2. For \((x^2+4y)^2\), the square terms are \(x^4\) and \(16y^2\), and the middle term is \(2(x^2)(4y)=8x^2y\). Thus the result is \(x^4+8x^2y+16y^2\).

1) \(\frac{4}{9}x^2-8x+36\) 2) \(x^4+8x^2y+16y^2\)

- Use the area formula for a square. - Apply the appropriate binomial-square identity to each decomposition. - Compare the mental arithmetic required by the two methods.

1. a) \((2+0.5)^2=2^2+2 \cdot 2 \cdot 0.5+(0.5)^2=4+2+0.25=6.25\), so the area is \(6.25\,\text{ft}^2\). 2. b) \((3-0.5)^2=3^2-2 \cdot 3 \cdot 0.5+(0.5)^2=9-3+0.25=6.25\), so the area is \(6.25\,\text{ft}^2\). 3. Both decompositions give the same area. A valid comparison may favor either method if the reasoning is clear; the two calculations have similar complexity.

a) \(6.25\,\text{ft}^2\) b) \(6.25\,\text{ft}^2\) c) Answers may vary; both methods are similarly efficient when justified.

- Identify the repeated expression in each pair of factors. - Rewrite each product in the form \((A+B)(A-B)\). - Expand any remaining binomial square and distribute a preceding negative sign carefully.

1. Group the repeated expression: \(((x+y)+3)((x+y)-3)=(x+y)^2-9=x^2+2xy+y^2-9\). 2. Write the product as \(((a-b)+c)((a-b)-c)=(a-b)^2-c^2=a^2-2ab+b^2-c^2\). 3. Write the product as \((7+(m+n))(7-(m+n))=49-(m+n)^2=49-m^2-2mn-n^2\).

1) \(x^2+2xy+y^2-9\) 2) \(a^2-2ab+b^2-c^2\) 3) \(49-m^2-2mn-n^2\)

- Group repeated sums or differences into a single expression. - Look for the structure \((A+B)(A-B)\). - When subtracting a binomial square, distribute the negative sign to every term.

1. a) Group the terms as \(((r+s)+(t+u))((r+s)-(t+u))\). Then \((r+s)^2-(t+u)^2\). 2. Expand both squares: \(r^2+2rs+s^2-(t^2+2tu+u^2)=r^2+2rs+s^2-t^2-2tu-u^2\). 3. b) Rewrite the factors as \((10-(x-2y))(10+(x-2y))\). Then the product is \(100-(x-2y)^2\). 4. Expand and simplify: \(100-(x^2-4xy+4y^2)=100-x^2+4xy-4y^2\).

a) \(r^2+2rs+s^2-t^2-2tu-u^2\) b) \(100-x^2+4xy-4y^2\)

- Check the signs when placing an expression after a minus sign. - Look for the form \((A-B)(A+B)\). - Expand the remaining binomial square and distribute the negative sign.

1. The second factor is \(r+(s-t)\). 2. The first factor is \(r-(s-t)\), since \(r-(s-t)=r-s+t\). 3. Apply the difference-of-squares identity: \((r-(s-t))(r+(s-t))=r^2-(s-t)^2\). 4. Expand and simplify: \(r^2-(s^2-2st+t^2)=r^2-s^2+2st-t^2\).

\(r^2-s^2+2st-t^2\)

- Look for products in the form \((A-B)(A+B)\). - In part a), simplify two factors at a time. - In part b), expand the square before combining like terms.

1. a) First, \((x-3)(x+3)=x^2-9\). 2. Then \((x^2-9)(x^2+9)=x^4-81\). 3. b) Expand the square: \((k+4)^2=k^2+8k+16\). 4. Subtract \(8k\): \(k^2+8k+16-8k=k^2+16\).

a) \(x^4-81\) b) \(k^2+16\)

- Work from left to right, combining two factors at a time. - Look for the same sum-and-difference pattern at every stage. - Apply the power-of-a-power rule when squaring a power.

1. Multiply the first two factors: \((y-2)(y+2)=y^2-4\). 2. Multiply by the third factor: \((y^2-4)(y^2+4)=y^4-16\). 3. Multiply by the fourth factor: \((y^4-16)(y^4+16)=y^8-256\).

\(y^8-256\)

- Group the nonmatching terms into one expression. - Rewrite each product in the form \((A+B)(A-B)\). - Distribute the negative sign after expanding the grouped square.

1. a) Group the factors as \((m+(2n+3p))(m-(2n+3p))\). This equals \(m^2-(2n+3p)^2\). 2. Expand and simplify: \(m^2-(4n^2+12np+9p^2)=m^2-4n^2-12np-9p^2\). 3. b) Group the factors as \((4u+(v-5w))(4u-(v-5w))\). This equals \(16u^2-(v-5w)^2\). 4. Expand and simplify: \(16u^2-(v^2-10vw+25w^2)=16u^2-v^2+10vw-25w^2\).

a) \(m^2-4n^2-12np-9p^2\) b) \(16u^2-v^2+10vw-25w^2\)

- Match each expression to a sum-square or difference-square identity. - Square coefficients and variable powers separately. - For each middle term, multiply twice the product of the two binomial terms.

1. \((0.5x^2+0.2y^3)^2=0.25x^4+0.2x^2y^3+0.04y^6\). 2. \(\left(\frac{3}{4}a-\frac{1}{3}b\right)^2=\frac{9}{16}a^2-\frac{1}{2}ab+\frac{1}{9}b^2\). 3. \((1.2p^2-0.5q)^2=1.44p^4-1.2p^2q+0.25q^2\).

1) \(0.25x^4+0.2x^2y^3+0.04y^6\) 2) \(\frac{9}{16}a^2-\frac{1}{2}ab+\frac{1}{9}b^2\) 3) \(1.44p^4-1.2p^2q+0.25q^2\)

- Identify \(a\) by comparing each number with \(10+a\). - Compare the complete binomial-square formula with the approximation. - The absolute error comes from the term that was omitted.

1. For \(10.2^2\), \(a=0.2\), so the estimate is \(100+20 \cdot 0.2=104\). For \(10.02^2\), \(a=0.02\), so the estimate is \(100+20 \cdot 0.02=100.4\). 2. The exact values are \((10+0.2)^2=100+4+0.04=104.04\) and \((10+0.02)^2=100+0.4+0.0004=100.4004\). 3. The absolute errors are \(|104.04-104|=0.04\) and \(|100.4004-100.4|=0.0004\). 4. The omitted term is exactly \(a^2\). Since \(0.02^2\) is much smaller than \(0.2^2\), the second error is much smaller.

1) \(104\) and \(100.4\) 2) \(104.04\) and \(100.4004\) 3) \(0.04\) and \(0.0004\) 4) The error equals the omitted term \(a^2\).

- Write \(0.99\) in the form \(1-a\). - Compare the full identity with the shortened approximation. - Recall the sign of the square of a nonzero real number.

1. Since \(0.99=1-0.01\), let \(a=0.01\). The approximation is \(1-2 \cdot 0.01=0.98\). The exact value is \((1-0.01)^2=1-0.02+0.0001=0.9801\). 2. The absolute error is \(|0.9801-0.98|=0.0001\). 3. The exact value minus the approximation is \((1-2a+a^2)-(1-2a)=a^2\). For real \(a\ne0\), \(a^2>0\), so the exact value is greater by \(a^2\).

1) Approximate value: \(0.98\); exact value: \(0.9801\) 2) \(0.0001\) 3) The exact value exceeds the approximation by \(a^2>0\).

- In the expansion method, distribute the subtraction sign across the entire second trinomial. - In the factoring method, treat each binomial as one complete term. - Compare the like terms produced by the two expansions.

1. Expanding gives \((x^2+10x+25)-(x^2-10x+25)=20x\). 2. Factoring gives \(((x+5)-(x-5))((x+5)+(x-5))=10\cdot2x=20x\). 3. In the expanded method, both squares contain the same \(x^2\)-term. Subtraction produces \(x^2-x^2=0\).

a) Both methods give \(20x\). b) The \(x^2\)-terms cancel because they have equal coefficients and are subtracted.

- In part a), distribute the subtraction sign across the second trinomial. - In part b), treat each binomial as one complete term. - For part c), identify the number midway between \(101\) and \(99\).

1. a) \((n^2+2n+1)-(n^2-2n+1)=4n\). 2. b) \(((n+1)-(n-1))((n+1)+(n-1))=2\cdot2n=4n\). 3. c) Here \(n=100\), so \(101^2-99^2=4\cdot100=400\).

a) \(4n\) b) \(4n\) c) \(400\)

- Expand each binomial square. - Write a number such as \(31\) as \(10a+1\). - Combine the two equal middle products when deriving the identity.

1. Expand: \((10a+1)^2=100a^2+20a+1\). 2. For \(31^2\), use \(a=3\): \(100\cdot3^2+20\cdot3+1=900+60+1=961\). For \(61^2\), use \(a=6\): \(100\cdot6^2+20\cdot6+1=3600+120+1=3721\). 3. \((10a+2)^2=100a^2+40a+4\). With \(a=3\), \(32^2=100\cdot9+40\cdot3+4=1024\).

1) \((10a+1)^2=100a^2+20a+1\) 2) \(31^2=961\); \(61^2=3721\) 3) \((10a+2)^2=100a^2+40a+4\); \(32^2=1024\)

- Use the nearest greater multiple of \(10\). - Write the number as \(10n-1\). - Apply the difference-square identity term by term.

1. a) \(19^2=(20-1)^2=400-40+1=361\), and \(39^2=(40-1)^2=1600-80+1=1521\). 2. b) \((10n-1)^2=100n^2-20n+1\). 3. c) For \(89^2\), use \(n=9\): \(100\cdot9^2-20\cdot9+1=8100-180+1=7921\).

a) \(19^2=361\); \(39^2=1521\) b) \((10n-1)^2=100n^2-20n+1\) c) \(89^2=7921\)

- Compare the trinomial with \(A^2-2AB+B^2\). - Recall that the square of a real number is nonnegative. - Find when the quantity inside the square equals \(0\).

1. a) \(T(0)=36\), \(T(6)=36-72+36=0\), and \(T(10)=100-120+36=16\). 2. b) \(x^2-12x+36=(x-6)^2\). 3. c) A square is always nonnegative, so \((x-6)^2\ge0\). Its minimum value is \(0\), reached when \(x-6=0\), or \(x=6\).

a) \(36\), \(0\), and \(16\) b) \(T=(x-6)^2\) c) The minimum value is \(0\) at \(x=6\), and \(T\) is never negative.

- Find the square roots of the outer terms. - A perfect-square trinomial has a middle term equal to twice the product of its binomial terms. - All three terms must fit the same binomial-square pattern.

1. The outer terms are \(\left(\frac{1}{4}m\right)^2\) and \(n^2\), and \(2\cdot\frac{1}{4}m\cdot n=\frac{1}{2}mn\). Thus the expression is \(\left(\frac{1}{4}m+n\right)\left(\frac{1}{4}m+n\right)\). 2. The binomial terms are \(0.2p\) and \(0.5q\). Their doubled product is \(0.2pq\), so either \(0.2pq\) or \(-0.2pq\) makes a binomial square. 3. If \(x^2+5x+6=(x+c)^2\), then \(2c=5\), so \(c=2.5\). The constant would have to be \(c^2=6.25\), not \(6\). Therefore, the trinomial is not a binomial square.

1) \(\left(\frac{1}{4}m+n\right)\left(\frac{1}{4}m+n\right)\) 2) \(0.2pq\) or \(-0.2pq\) 3) No. A middle term of \(5x\) would require a constant term of \(6.25\), not \(6\).

- Expand both binomial squares. - Combine corresponding like terms. - Compare the signs of the two mixed-product terms.

1. Expand the sum: \((a+b)^2+(a-b)^2=(a^2+2ab+b^2)+(a^2-2ab+b^2)\). 2. Combine like terms: \(2a^2+2b^2=2(a^2+b^2)\). 3. The terms \(2ab\) and \(-2ab\) cancel, so the final expression contains no \(ab\)-term.

\((a+b)^2+(a-b)^2=2a^2+2b^2=2(a^2+b^2)\). The \(ab\)-terms cancel.

- Expand each binomial square separately. - Combine like terms and notice which linear terms cancel. - Factor the final polynomial.

1. Expand both squares: \((2x+3)^2=4x^2+12x+9\) and \((3x-2)^2=9x^2-12x+4\). 2. Add and combine like terms: \(4x^2+12x+9+9x^2-12x+4=13x^2+13\). 3. Factor out \(13\): \(13x^2+13=13(x^2+1)\). Therefore, the identity is true.

Both sides simplify to \(13x^2+13\), so the identity is true for every real \(x\).

- A binomial square has three terms after expansion. - Identify the missing middle product in Tim’s work. - For \(1005\), use a nearby multiple of \(1000\).

1. a) Tim omitted the middle term \(-2AB\) from \((A-B)^2=A^2-2AB+B^2\). 2. b) \((40-1)^2=40^2-2\cdot40\cdot1+1^2=1600-80+1=1521\). 3. c) \((1000+5)^2=1000^2+2\cdot1000\cdot5+5^2=1{,}000{,}000+10{,}000+25=1{,}010{,}025\).

a) He omitted the middle term \(-2AB\). b) \(1521\) c) \(1{,}010{,}025\)

- What perfect-square trinomial begins with \(x^2 + 4x\)? - How does the constant inside the parentheses relate to \(c\)? - What is the vertex of a quadratic written as \(a(x - h)^2\)?

1. Factor \(2\): \(f(x) = 2\left(x^2 + 4x + \frac{c}{2}\right)\). 2. The trinomial inside the parentheses must be \((x + 2)^2 = x^2 + 4x + 4\). Therefore, \(\frac{c}{2} = 4\), so \(c = 8\). 3. Then \(f(x) = 2(x + 2)^2\), whose vertex is \((-2, 0)\).

1) \(c = 8\) 2) \((-2, 0)\)

- Compare each expression with a binomial-square pattern. - Use the middle term to determine a missing binomial term. - The first and last terms of the expansion are squares.

1. a) The middle term is \(2(x)(6) = 12x\), and \(6^2 = 36\). Thus, \((x + 6)^2 = x^2 + 12x + 36\). 2. b) Since the first square is \(25x^2\), the first term is \(5x\). The middle term is \(2(5x)(4y) = 40xy\). Thus, \((5x - 4y)^2 = 25x^2 - 40xy + 16y^2\). 3. c) The middle term requires \(2(u)(0.1) = 0.4a\), so \(u = 2a\). Then \(u^2 = 4a^2\). Thus, \((2a + 0.1)^2 = 4a^2 + 0.4a + 0.01\).

a) \((x + 6)^2 = x^2 + 12x + 36\) b) \((5x - 4y)^2 = 25x^2 - 40xy + 16y^2\) c) \((2a + 0.1)^2 = 4a^2 + 0.4a + 0.01\)

- What must the vertical shift be in vertex form when the vertex lies on the \(x\)-axis? - Then \(x^2 + kx + 16\) must be a perfect-square trinomial. - Which two binomial squares have constant term \(16\)?

1. A quadratic with leading coefficient \(1\) and a vertex on the \(x\)-axis must be a perfect square: \(p(x) = (x - h)^2\). 2. Since the constant term is \(16\), the squared binomial must use \(4\): \((x + 4)^2 = x^2 + 8x + 16\) or \((x - 4)^2 = x^2 - 8x + 16\). 3. Therefore, \(k = 8\) gives \(p(x) = (x + 4)^2\), whose vertex is \((-4, 0)\). 4. The value \(k = -8\) gives \(p(x) = (x - 4)^2\), whose vertex is \((4, 0)\).

a) \(k = 8\) or \(k = -8\) b) For \(k = 8\): \((-4, 0)\); for \(k = -8\): \((4, 0)\)

- Translate the perimeter and area into equations involving \(a+b\) and \(ab\). - Use the identity \((a+b)^2=a^2+2ab+b^2\). - Relate \(a^2+b^2\) to the square of the diagonal.

1. If the side lengths are \(a\) and \(b\), then \(a+b=14\) and \(ab=48\). 2. Use \((a+b)^2=a^2+2ab+b^2\): \(14^2=a^2+b^2+2\cdot 48\). 3. Since \(d^2=a^2+b^2\), \(196=d^2+96\), so \(d^2=100\). 4. Therefore, \(d=10\,\text{cm}\).

\(d=10\,\text{cm}\)

- Match each expression to a binomial identity. - Use the doubled product to determine missing terms. - Check the first and last terms as perfect squares.

1. a) The middle term \(12x\) is \(2(x)(6)\), so the completed identity is \(x^2 + 12x + 36 = (x + 6)^2\). 2. b) Use the difference of squares: \((4a - 7)(4a + 7) = 16a^2 - 49\). 3. c) Since \(25 = 5^2\) and \(-20y = -2(5)(2y)\), the completed identity is \(25 - 20y + 4y^2 = (5 - 2y)^2\).

a) \(x^2 + 12x + 36 = (x + 6)^2\) b) \(16a^2 - 49 = (4a - 7)(4a + 7)\) c) \(25 - 20y + 4y^2 = (5 - 2y)^2\)

- Match the leading term to \(a^2x^2\). - Use the middle term \(2abx\) to find \(b\). - The constant term is \(b^2\).

1. Compare with \((ax + b)^2 = a^2x^2 + 2abx + b^2\). 2. Since \(a^2 = 4\) and \(a > 0\), \(a = 2\). 3. The middle term gives \(2(2)b = 12\), so \(b = 3\). 4. Therefore, \(c = b^2 = 9\), and the expression factors as \((2x + 3)^2\).

\(c = 9\); \((2x + 3)^2\)

- In the identity, examine how \(a\) and \(b\) determine the middle coefficient and constant term. - For the check, evaluate the original and expanded forms separately. - In part c), find a number that adds to \(3\) to make \(10\) and multiplies by \(3\) to make \(21\).

1. For a), \(a = 12\) and \(b = -3\), so \((x + 12)(x - 3) = x^2 + 9x - 36\). 2. For b), \((x - 5)(x - 4) = x^2 - 9x + 20\). At \(x = 10\), the original product is \((10 - 5)(10 - 4) = 30\), and the expanded form is \(10^2 - 9 \cdot 10 + 20 = 30\). 3. For c), the middle coefficient requires \(3 + c = 10\), and the constant term requires \(3c = 21\). Both give \(c = 7\).

a) \(x^2 + 9x - 36\) b) \(x^2 - 9x + 20\); the check gives \(30 = 30\) c) \(c = 7\), because \(3 + 7 = 10\) and \(3 \cdot 7 = 21\)

- Interpret the square as a product of two identical binomials. - Identify the common factor in the first two terms. - Multiplying by \(100\) shifts a whole number two places to the left.

1. a) \((10x + 5)^2 = 100x^2 + 100x + 25\). 2. b) Factor the first two terms: \(100x(x + 1) + 25\). 3. c) For \(35\), \(x = 3\). Then \(x(x + 1) = 3 \cdot 4 = 12\), so \(100 \cdot 12 + 25 = 1225\).

a) \(100x^2 + 100x + 25\) b) \(100x(x + 1) + 25\) c) For \(x = 3\), \(3 \cdot 4 = 12\), and \(100 \cdot 12 + 25 = 1225\).

- Match the first and last terms to perfect squares. - Use the middle term to confirm the two binomial terms. - Remember that the middle term is twice the product.

1. a) Since \(49 = 7^2\) and \(2(a)(7) = 14a\), the missing term is \(7\). 2. b) Since \(9x^2 = (3x)^2\) and \(-2(3x)(4) = -24x\), the missing term is \(3x\). 3. c) Since \(25 = 5^2\), the missing binomial term is \(5\), and the first term of the expansion is \((2y)^2 = 4y^2\).

a) \((a + 7)^2 = a^2 + 14a + 49\) b) \((3x - 4)^2 = 9x^2 - 24x + 16\) c) \((2y + 5)^2 = 4y^2 + 20y + 25\)

- Match the expression to \(A^2 \pm 2AB + B^2\). - Use the square terms to identify the binomial terms. - Use the middle term to verify your choices.

1. The first term must be \(x\), since \(x^2\) is the first square and \(2(x)(3) = 6x\). 2. The missing term is \(\frac{1}{2}b\), since \(2(2a)\left(\frac{1}{2}b\right) = 2ab\). 3. The binomial term is \(n\), and the first square is \(\left(\frac{1}{3}m\right)^2 = \frac{1}{9}m^2\).

1) \((x + 3)^2 = x^2 + 6x + 9\) 2) \(\left(2a - \frac{1}{2}b\right)^2 = 4a^2 - 2ab + \frac{1}{4}b^2\) 3) \(\left(\frac{1}{3}m + n\right)^2 = \frac{1}{9}m^2 + \frac{2}{3}mn + n^2\)

- Identify the square roots of the first and last terms. - The middle term is twice the product of the binomial terms. - Use the sign in the binomial to determine the middle sign.

1. a) Since \(25 = 5^2\), the missing binomial term is \(5\). The middle term is \(2(3x)(5) = 30x\). 2. b) Since \(0.16x^2 = (0.4x)^2\), the missing first term is \(0.4x\). The middle term is \(2(0.4x)(0.2y^2) = 0.16xy^2\).

a) \((3x + 5)^2 = 9x^2 + 30x + 25\) b) \((0.4x - 0.2y^2)^2 = 0.16x^2 - 0.16xy^2 + 0.04y^4\)

- Find the square roots of the first and last terms. - Check that twice their product matches the middle term. - Compare the sum-square and difference-square patterns.

1. The first term is \(\left(\frac{1}{2}a^3\right)^2\), and the last term is \((3b^2)^2\). 2. Their doubled product is \(2\left(\frac{1}{2}a^3\right)(3b^2) = 3a^3b^2\), so the original binomial is \(\left(\frac{1}{2}a^3 + 3b^2\right)^2\). 3. To make the middle term negative, replace the plus sign with a minus sign: \(\left(\frac{1}{2}a^3 - 3b^2\right)^2\).

a) \(\left(\frac{1}{2}a^3 + 3b^2\right)^2\) b) \(\left(\frac{1}{2}a^3 - 3b^2\right)^2\); changing the sign changes only the middle term’s sign.

- Match each expression to \(A^2\pm2AB+B^2\). - Use the first and last square terms to identify the binomial terms. - For part 3, expand both sides and compare the results.

1. In \((A+B)^2=A^2+2AB+B^2\), let \(A=4x\). Since \(2(4x)B=40xy\), \(B=5y\). Thus \((4x+5y)^2=16x^2+40xy+25y^2\). 2. Since \(25a^2=(5a)^2\), \(1=1^2\), and \(-2(5a)(1)=-10a\), the expression is \((5a-1)^2\). 3. Expanding gives \((x-7)^2=x^2-14x+49\) and \((7-x)^2=49-14x+x^2=x^2-14x+49\). Therefore the expressions are equal for every real \(x\).

1) \((4x+5y)^2=16x^2+40xy+25y^2\) 2) \((5a-1)^2\) 3) Both sides simplify to \(x^2-14x+49\).

- Match each statement to a binomial identity. - Use square roots of the first and last terms to identify missing factors. - Check the middle term in part b) using twice the product.

1. a) Since \(16x^2=(4x)^2\) and \((3y)^2=9y^2\), \((4x-3y)(4x+3y)=16x^2-9y^2\). 2. b) Since \(9=3^2\) and \(2(5a)(3)=30a\), \((5a+3)^2=25a^2+30a+9\). 3. c) Since \(0.25z^2=(0.5z)^2\), \((4-0.5z)(4+0.5z)=16-0.25z^2\).

a) \((4x-3y)(4x+3y)=16x^2-9y^2\) b) \((5a+3)^2=25a^2+30a+9\) c) \((4-0.5z)(4+0.5z)=16-0.25z^2\)

- Compare each statement with \(A^2-B^2=(A-B)(A+B)\). - Take square roots of the visible square terms. - Check decimal place value when squaring \(0.1\).

1. a) Since \(4k^2=(2k)^2\) and \(5^2=25\), \((2k-5)(2k+5)=4k^2-25\). 2. b) Since \(100x^2=(10x)^2\) and \((9y)^2=81y^2\), \(100x^2-81y^2=(10x-9y)(10x+9y)\). 3. c) Since \((0.1z)^2=0.01z^2\) and \(1=1^2\), \(0.01z^2-1=(0.1z-1)(0.1z+1)\).

a) \((2k-5)(2k+5)=4k^2-25\) b) \(100x^2-81y^2=(10x-9y)(10x+9y)\) c) \(0.01z^2-1=(0.1z-1)(0.1z+1)\)

- Find the midpoint of each pair of factors. - Find each factor’s distance from that midpoint. - Compare \(n^2-k^2\) with \(n^2\).

1. a) \(21\cdot19=(20+1)(20-1)=400-1=399\). Also, \(52\cdot48=(50+2)(50-2)=2500-4=2496\). 2. b) \((n-k)(n+k)=n^2-k^2\). For real \(k\ne0\), \(k^2>0\), so the product is exactly \(k^2\) less than \(n^2\). 3. c) \(35\cdot45=(40-5)(40+5)=40^2-5^2=1600-25=1575\). In general, factors equally spaced by \(k\) from midpoint \(n\) have product \(n^2-k^2\).

a) \(21\cdot19=399\); \(52\cdot48=2496\) b) The product is \(k^2\) less than \(n^2\). c) \(35\cdot45=1575\); the shortcut follows from \((n-k)(n+k)=n^2-k^2\).

- Express the middle term as twice a product. - Use the square roots of known outer terms. - Expand your completed binomial square to verify it.

1. a) Since \(10x=2\cdot x\cdot5\), the missing term is \(25\): \(x^2+10x+25=(x+5)^2\). 2. b) The outer terms are \((2a)^2\) and \(3^2\), so the middle term is \(2\cdot2a\cdot3=12a\): \(4a^2-12a+9=(2a-3)^2\). 3. c) Since \(16=4^2\) and \(24y=2\cdot3y\cdot4\), the missing first term is \(9y^2\): \(9y^2+24y+16=(3y+4)^2\).

a) \(x^2+10x+25=(x+5)^2\) b) \(4a^2-12a+9=(2a-3)^2\) c) \(9y^2+24y+16=(3y+4)^2\)

- Group the terms into two sum-and-difference pairs. - Use \((u+v)^2+(u-v)^2=2u^2+2v^2\). - Simplify the remaining difference of two binomial squares.

1. Group the first pair using \(u=a+b\): \((u+c)^2+(u-c)^2=2u^2+2c^2=2(a+b)^2+2c^2\). 2. Group the second pair using \(v=a-b\): \((v+c)^2+(v-c)^2=2v^2+2c^2=2(a-b)^2+2c^2\). 3. Subtract the pairs: \(2(a+b)^2-2(a-b)^2=2\left((a+b)^2-(a-b)^2\right)\). 4. Since \((a+b)^2-(a-b)^2=4ab\), the expression equals \(2\cdot4ab=8ab\).

The left side simplifies to \(8ab\), so the identity is true.

- The middle term is twice the product of the binomial terms. - Use square roots of the known square terms. - Compare each identity with \(A^2\pm2AB+B^2\).

1. a) Since \(30a^2b^3=2\cdot3a^2\cdot5b^3\), the missing binomial term is \(5b^3\). Its square is \(25b^6\). 2. b) Since \(4x^6=(2x^3)^2\), the missing first term is \(2x^3\). The final square term is \(\left(\frac{1}{2}y^2\right)^2=\frac{1}{4}y^4\).

a) \((3a^2+5b^3)^2=9a^4+30a^2b^3+25b^6\) b) \(\left(2x^3-\frac{1}{2}y^2\right)^2=4x^6-2x^3y^2+\frac{1}{4}y^4\)