6.4 Special products  (Page 4/5)

${(w+4)}^2$

${(q+12)}^2$

${\left(y+\frac{1}{4}\right)}^2$

${\left(x+\frac{2}{3}\right)}^2$

${(b-7)}^2$

${(y-6)}^2$

${\left(m-15\right)}^2$

${\left(p-13\right)}^2$

${(3d+1)}^2$

${\left(4a+10\right)}^2$

${(2q+\frac{1}{3})}^2$

${\left(3z+\frac{1}{5}\right)}^2$

${\left(3x-y\right)}^2$

${\left(2y-3z\right)}^2$

${\left(\frac{1}{5}x-\frac{1}{7}y\right)}^2$

${\left(\frac{1}{8}x-\frac{1}{9}y\right)}^2$

${(3x^2+2)}^2$

${(5u^2+9)}^2$

${(4y^3-2)}^2$

${(8p^3-3)}^2$

$(m-7)(m+7)$

$(c-5)(c+5)$

$\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)$

$\left(b+\frac{6}{7}\right)\left(b-\frac{6}{7}\right)$

$(5k+6)(5k-6)$

$(8j+4)(8j-4)$

$(11k+4)(11k-4)$

$(9c+5)(9c-5)$

$(11-b)(11+b)$

$(13-q)(13+q)$

$(5-3x)(5+3x)$

$(4-6y)(4+6y)$

$(9c-2d)(9c+2d)$

$(7w+10x)(7w-10x)$

$\left(m+\frac{2}{3}n\right)\left(m-\frac{2}{3}n\right)$

$\left(p+\frac{4}{5}q\right)\left(p-\frac{4}{5}q\right)$

$(ab-4)(ab+4)$

$(xy-9)(xy+9)$

$(uv-\frac{3}{5})(uv+\frac{3}{5})$

$(rs-\frac{2}{7})(rs+\frac{2}{7})$

$(2x^2-3y^4)(2x^2+3y^4)$

$(6m^3-4n^5)(6m^3+4n^5)$

$(12p^3-11q^2)(12p^3+11q^2)$

$(15m^2-8n^4)(15m^2+8n^4)$

ⓐ $\left(p-3\right)\left(p+3\right)$
ⓑ ${\left(t-9\right)}^2$
ⓒ ${\left(m+n\right)}^2$
ⓓ $(2x+y)(x-2y)$

1. ⓐ ${(2r+12)}^2$
2. ⓑ $(3p+8)(3p-8)$
3. ⓒ $(7a+b)(a-7b)$
4. ⓓ ${\left(k-6\right)}^2$

1. ⓐ ${(a^5-7b)}^2$
2. ⓑ $(x^2+8y)(8x-y^2)$
3. ⓒ $(r^6+s^6)(r^6-s^6)$
4. ⓓ ${(y^4+2z)}^2$

1. ⓐ $(x^5+y^5)(x^5-y^5)$
2. ⓑ ${(m^3-8n)}^2$
3. ⓒ ${(9p+8q)}^2$
4. ⓓ $(r^2-s^3)(r^3+s^2)$

Mental math You can use the product of conjugates pattern to multiply numbers without a calculator. Say you need to multiply 47 times 53. Think of 47 as $50-3$ and 53 as $50+3$ .

1. ⓐ Multiply $(50-3)(50+3)$ by using the product of conjugates pattern, $\left(a-b\right)\left(a+b\right)=a^2-b^2$ .
2. ⓑ Multiply $47·53$ without using a calculator.
3. ⓒ Which way is easier for you? Why?

Mental math You can use the binomial squares pattern to multiply numbers without a calculator. Say you need to square 65. Think of 65 as $60+5$ .

1. ⓐ Multiply ${(60+5)}^2$ by using the binomial squares pattern, ${\left(a+b\right)}^2=a^2+2ab+b^2$ .
2. ⓑ Square 65 without using a calculator.
3. ⓒ Which way is easier for you? Why?

How do you decide which pattern to use?

Why does ${(a+b)}^2$ result in a trinomial, but $(a-b)(a+b)$ result in a binomial?

Marta did the following work on her homework paper:

Explain what is wrong with Marta’s work.

Use the order of operations to show that ${\left(3+5\right)}^2$ is 64, and then use that numerical example to explain why ${\left(a+b\right)}^2\ne a^2+b^2$ .