- Check whether the first and last terms are squares. - For the three-term expression, compare the middle term with twice the product of the square roots of the outer terms. - For the two-term expression, notice the subtraction between two squares.

1. For part a, \(x^2+6x+9=x^2+2\cdot x\cdot3+3^2\), so it is a perfect-square trinomial and equals \((x+3)^2\). 2. For part b, \(x^2-9=x^2-3^2\), so it is a difference of squares and equals \((x-3)(x+3)\).

a) Perfect-square trinomial: \((x+3)^2\) b) Difference of squares: \((x-3)(x+3)\)

- Focus only on the middle term \(2AB\). - Substitute the given expressions for \(A\) and \(B\).

1. Substitute \(A=x\) and \(B=2\) into the middle term \(2AB\). 2. \(2AB=2\cdot x\cdot2=4x\).

The \(2AB\) term becomes \(4x\).

- Write both terms as squares. - Think of two conjugate factors whose product has no middle term.

1. Recognize \(16=4^2\), so the expression is \(x^2-4^2\). 2. A difference of two squares factors as \((x-4)(x+4)\).

\((x-4)(x+4)\)

- Rewrite the square as a product of two identical binomials. - Identify the products that contribute to the \(x\)-term, then compare its coefficient with the blank.

1. The middle term of \((x+5)^2\) is \(2\cdot x\cdot5=10x\). 2. Therefore, the missing coefficient is \(10\).

\(10\)

- Look at the highest power of the variable. - Keep each sign with its coefficient.

1. The largest exponent of \(x\) is \(2\), so the expression has degree \(2\). 2. The coefficient of \(x^2\) is \(3\), the coefficient of \(x\) is \(-5\), and the constant term is \(2\).

Degree \(2\); \(x^2\)-coefficient \(3\); \(x\)-coefficient \(-5\); constant \(2\).

- A quadratic expression has degree \(2\) after simplification. - An expression does not need to be expanded to be quadratic. - Consider the leading power produced by the product in part c).

1. Expression a) has highest power \(x^2\), so it is quadratic. 2. Expression b) has highest power \(x^4\), so it is not quadratic. 3. Expression c) equals \(x^2-1\), so it is quadratic even though it is given in product form.

a) Quadratic b) Not quadratic c) Quadratic

- Compare what you can tell immediately from each form. - Think about whether multiplying a real number by itself can produce a negative value.

1. The form \((x-3)^2\) shows a real number multiplied by itself. 2. A square is never negative, so \((x-3)^2\ge 0\) for every real \(x\).

\((x-3)^2\), because the square of a real number cannot be negative.

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

- Find the midpoint of each pair of factors. - Express the two factors as equal distances above and below that midpoint. - Keep the difference-of-squares rewrite in your answer before evaluating.

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

- Rewrite each difference of squares as a product before doing arithmetic. - Subtract once and add once. - Pay attention to the sign of the first factor in part b).

1. a) \((45-35)(45+35)=10\cdot80=800\). 2. b) \((16-24)(16+24)=-8\cdot40=-320\). 3. c) \((101-99)(101+99)=2\cdot200=400\).

a) \((45-35)(45+35)=800\) b) \((16-24)(16+24)=-320\) c) \((101-99)(101+99)=400\)

- 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 benchmark that makes the arithmetic easy. - Write the number as a sum or difference from that benchmark. - Keep the identity expansion visible before evaluating the arithmetic.

1. a) \(21=20+1\), so \((20+1)^2=20^2+2\cdot20\cdot1+1^2=441\). 2. b) \(49=50-1\), so \((50-1)^2=50^2-2\cdot50\cdot1+1^2=2401\). 3. c) \(10.1=10+0.1\), so \((10+0.1)^2=10^2+2\cdot10\cdot0.1+(0.1)^2=102.01\). 4. d) \(199=200-1\), so \((200-1)^2=200^2-2\cdot200\cdot1+1^2=39{,}601\).

a) \(21=20+1\): \((20+1)^2=20^2+2\cdot20\cdot1+1^2=441\) b) \(49=50-1\): \((50-1)^2=50^2-2\cdot50\cdot1+1^2=2401\) c) \(10.1=10+0.1\): \((10+0.1)^2=10^2+2\cdot10\cdot0.1+(0.1)^2=102.01\) d) \(199=200-1\): \((200-1)^2=200^2-2\cdot200\cdot1+1^2=39{,}601\)

- 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 every difference of squares as a product first. - One factor is the difference and the other is the sum. - Look for a resulting factor that is 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) \(\left(5\frac{1}{2}-4\frac{1}{2}\right)\left(5\frac{1}{2}+4\frac{1}{2}\right)=10\) b) \((205-5)(205+5)=42{,}000\) c) \((8.7-1.3)(8.7+1.3)=74\)

- Factor each expression as \((u-v)(u+v)\). - Keep the factorization visible before multiplying. - Compare the two resulting products.

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

\(A=(55-45)(55+45)=1000\) \(B=(102.5-97.5)(102.5+97.5)=1000\) Therefore, \(A=B\).

- 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 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) \((3x)^2+2(3x)(4)+4^2=9x^2+24x+16\) b) \(7^2-2(7)(2b)+(2b)^2=49-28b+4b^2\) c) \((5k)^2-1^2=25k^2-1\)

- Check whether the first and last terms are perfect squares. - For a three-term expression, compare the middle term with twice the product of the square roots of the outer terms. - For two square terms separated by subtraction, think about the product of conjugates that would expand back to the expression.

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

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

- Each small rectangle's area is the product of its two side lengths. - Add the four smaller areas. - Find the full side length from the labeled parts of an edge.

1. The four rectangle areas are \(x^2\), \(2x\), \(2x\), and \(4\). 2. Their total is \(x^2+2x+2x+4=x^2+4x+4\). 3. The whole square has side length \(x+2\), so its area is also \((x+2)^2\). 4. Therefore \((x+2)^2=x^2+4x+4\).

Rectangle areas: \(x^2\), \(2x\), \(2x\), \(4\) Identity: \((x+2)^2=x^2+4x+4\)

- Think about which term remains when \(x=0\). - Can you determine the sign of a product from the signs of its factors? - Two equal expressions can make different features easier to notice.

1. The expanded form has constant term \(-6\), giving value \(-6\) at \(x=0\). 2. The factored form shows the signs directly. At \(x=1\), the factors are \(4\) and \(-1\), so the value is negative. At \(x=4\), the factors are \(7\) and \(2\), so the value is positive. 3. The expanded and factored forms reveal different useful information without changing the expression.

a) Expanded form; value \(-6\). b) Factored form; negative at \(x=1\), positive at \(x=4\).

- The area of a small rectangle is its width times its height. - Use the known corner area to find the missing side piece first. - Add the pieces along each outside edge to get the full side lengths.

1. The upper-right rectangle has area \(12\) and width \(3\), so its height is \(12\div3=4\). 2. The full side lengths are \(x+3\) and \(x+4\). 3. The full area is \((x+3)(x+4)=x^2+3x+4x+12=x^2+7x+12\).

Unknown side length \(4\); area \(x^2+7x+12\).

- Find each square's area from its labeled side length. - The remaining area is what belongs to the large square but not the small one. - Combine like terms only after finding the difference.

1. The larger square has area \((x+2)^2\); the smaller has area \(x^2\). 2. The border area is \((x+2)^2-x^2\). 3. Expanding and subtracting gives \(x^2+4x+4-x^2=4x+4\).

\((x+2)^2-x^2=4x+4\)

- Which part of the expression can never be negative? - Consider how its size affects the whole expression. - Check whether your proposed smallest value can actually occur.

1. The square \((x+2)^2\) is never negative, so \(3(x+2)^2\ge0\). 2. Its smallest value is \(0\), which occurs when \(x+2=0\), or \(x=-2\). 3. Subtracting \(5\) gives a minimum value of \(-5\).

Minimum \(-5\), when \(x=-2\).

- Write the squared expression as a product of two identical binomials. - Try \(x=0\). Does Mia's equation hold for that value? - Multiply the two binomials term by term to find an expansion that always works.

1. Squaring \((x-4)\) means multiplying \((x-4)(x-4)\), not taking a difference of two squares. 2. Multiplication gives \(x^2-4x-4x+16=x^2-8x+16\). 3. Mia omitted the two mixed products and used the wrong sign for the constant term.

\((x-4)^2=x^2-8x+16\). The original answer misses the mixed term and incorrectly treats a squared difference as a difference of squares.

- 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.

- 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. The middle term is \(2\cdot x\cdot6=12x\), and \(6^2=36\). Thus, \((x+6)^2=x^2+12x+36\). 2. Since the first square is \(25x^2\), the first term is \(5x\). The middle term is \(2\cdot5x\cdot4y=40xy\). Thus, \((5x-4y)^2=25x^2-40xy+16y^2\). 3. The middle term requires \(2\cdot u\cdot0.1=0.4a\), 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) \((2a+0.1)^2=4a^2+0.4a+0.01\)

- 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) \((7z + 9)(7z - 9)\)

- 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. The middle term \(12x\) is \(2\cdot x\cdot6\), so the completed identity is \(x^2+12x+36=(x+6)^2\). 2. Use the difference of squares: \((4a-7)(4a+7)=16a^2-49\). 3. Since \(25=5^2\) and \(-20y=-2\cdot5\cdot2y\), 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) \(25-20y+4y^2=(5-2y)^2\)

- 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) \((s+5)(s-5)=s^2-25\) The square has \(25\,\text{ft}^2\) more area.

- Substitute the given test value into both sides first. - Compare the complete square of a binomial with the proposed two-term expression. - After expanding correctly, determine when the missing contribution becomes zero.

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

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

- Expand the squared binomial before combining the other terms. - Look for two squared quantities after taking out the outside factor. - Try substituting the given input into each form. Which form needs the least calculation?

1. Expand the square: \(Q(x)=2(x^2-6x+9)-50=2x^2-12x-32\). 2. First write \(Q(x)=2((x-3)^2-25)\). The difference of squares gives \(Q(x)=2((x-3)-5)((x-3)+5)=2(x-8)(x+2)\). 3. The factored form shows that \(x=8\) makes the factor \(x-8\) zero, so \(Q(8)=0\).

a) \(2x^2-12x-32\) b) \(2(x-8)(x+2)\) c) The factored form, because \(x=8\) makes one factor zero.

- 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 \(3 \cdot 7 = 21\)

- 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. Since \(49=7^2\) and \(2\cdot a\cdot7=14a\), the missing term is \(7\). 2. Since \(9x^2=(3x)^2\) and \(-2\cdot3x\cdot4=-24x\), the missing term is \(3x\). 3. 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 \(A^2\pm2AB+B^2\). - 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 \(2\cdot x\cdot3=6x\). 2. The missing term is \(\frac{1}{2}b\), since \(2\cdot2a\cdot\frac{1}{2}b=2ab\). 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) \(\left(\frac{1}{3}m+n\right)^2=\frac{1}{9}m^2+\frac{2}{3}mn+n^2\)

- 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) \(((x+y)+3)((x+y)-3)=(x+y)^2-9=x^2+2xy+y^2-9\) 2) \(((a-b)+c)((a-b)-c)=(a-b)^2-c^2=a^2-2ab+b^2-c^2\) 3) \((7+(m+n))(7-(m+n))=49-(m+n)^2=49-m^2-2mn-n^2\)

- 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) \(((r+s)+(t+u))((r+s)-(t+u))=(r+s)^2-(t+u)^2\) \(=r^2+2rs+s^2-t^2-2tu-u^2\) b) \((10-(x-2y))(10+(x-2y))=100-(x-2y)^2\) \(=100-x^2+4xy-4y^2\)

- 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: \(r^2-(s^2-2st+t^2)=r^2-s^2+2st-t^2\).

1. \(r+(s-t)\) 2. \(r-(s-t)\) 3. \(r^2-(s-t)^2=r^2-s^2+2st-t^2\)

- 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) \((m+(2n+3p))(m-(2n+3p))=m^2-(2n+3p)^2\) \(=m^2-4n^2-12np-9p^2\) b) \((4u+(v-5w))(4u-(v-5w))=16u^2-(v-5w)^2\) \(=16u^2-v^2+10vw-25w^2\)

- 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 \(a^2>0\).

- 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. Since \(16x^2=(4x)^2\) and \((3y)^2=9y^2\), \((4x-3y)(4x+3y)=16x^2-9y^2\). 2. Since \(9=3^2\) and \(2\cdot5a\cdot3=30a\), \((5a+3)^2=25a^2+30a+9\). 3. 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. Since \(4k^2=(2k)^2\) and \(5^2=25\), \((2k-5)(2k+5)=4k^2-25\). 2. Since \(100x^2=(10x)^2\) and \((9y)^2=81y^2\), \(100x^2-81y^2=(10x-9y)(10x+9y)\). 3. 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) \(0.01z^2-1=(0.1z-1)(0.1z+1)\)

- 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) Expansion method: \((x^2+10x+25)-(x^2-10x+25)=20x\) Difference-of-squares method: \(((x+5)-(x-5))((x+5)+(x-5))=10(2x)=20x\) b) The \(x^2\)-terms disappear because the two expanded squares contain equal \(x^2\)-terms that 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) \((n^2+2n+1)-(n^2-2n+1)=4n\) b) \(((n+1)-(n-1))((n+1)+(n-1))=2(2n)=4n\) c) With \(n=100\), \(101^2-99^2=4(100)=400\)

- 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. \((10a+1)^2=(10a)^2+2(10a)(1)+1^2=100a^2+20a+1\), proving the identity. 2. \(31^2=100(3)^2+20(3)+1=961\) \(61^2=100(6)^2+20(6)+1=3721\) 3. \((10a+2)^2=(10a)^2+2(10a)(2)+2^2=100a^2+40a+4\) \(32^2=100(3)^2+40(3)+4=1024\)

- 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.

Expand the left side: \((2x+3)^2+(3x-2)^2=(4x^2+12x+9)+(9x^2-12x+4)=13x^2+13=13(x^2+1)\). Therefore, the left side equals the right side 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) \(1{,}010{,}025\)

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

\(a^2+b^2=100\,\text{cm}^2\) \(d=10\,\text{cm}\)

- 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 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 \(2\cdot4x\cdot B=40xy\), \(B=5y\). Thus, \((4x+5y)^2=16x^2+40xy+25y^2\). 2. Since \(25a^2=(5a)^2\), \(1=1^2\), and \(-2\cdot5a\cdot1=-10a\), 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 \(x^2-14x+49\).

- 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 \((n-k)(n+k)=n^2-k^2\).

- Represent the two side lengths with variables. - Translate the area difference into a difference of squares. - Use the given sum to find the difference of the side lengths. - Solve the resulting pair of linear equations.

1. Let \(x\) be the larger side length and \(y\) the smaller side length. Then \(x+y=22\) and \(x^2-y^2=88\). 2. Factor the area difference: \((x-y)(x+y)=88\). 3. Since \(x+y=22\), \(22(x-y)=88\), so \(x-y=4\). 4. Solve the linear system \(x+y=22\) and \(x-y=4\) by adding the equations: \(2x=26\), so \(x=13\). Then \(y=9\).

The side lengths are \(13\,\text{cm}\) and \(9\,\text{cm}\).

- 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. Since \(10x=2\cdot x\cdot5\), the missing term is \(25\): \(x^2+10x+25=(x+5)^2\). 2. 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. 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) \(9y^2+24y+16=(3y+4)^2\)

- Look for pairs of expressions that differ only in the sign of one term. - Can you reduce the four squares to two simpler expressions first? - Which terms disappear when the two grouped results are subtracted?

1. Group the first two squares using \(u=a+b\): \((u+c)^2+(u-c)^2=2u^2+2c^2=2(a+b)^2+2c^2\). 2. Group the last two squares using \(v=a-b\): \((v+c)^2+(v-c)^2=2v^2+2c^2=2(a-b)^2+2c^2\). 3. Subtract these results: \(2(a+b)^2-2(a-b)^2=2\big((a+b)^2-(a-b)^2\big)\). 4. Since \((a+b)^2-(a-b)^2=4ab\), the left side equals \(8ab\).

Grouping pairs gives \(2(a+b)^2+2c^2-2(a-b)^2-2c^2\). This simplifies to \(2\big((a+b)^2-(a-b)^2\big)=2(4ab)=8ab\).