Factor Difference of Two Squares 3

Question 1 of 4

Factor.

\frac{1}{4} y^{2} - 25 x^{2}

$1/4y^2-25x^2$

1. $(\frac{1}{2} y + 5 x) (\frac{1}{2} y - 5 x)$ $(1/2y+5x)(1/2y-5x)$
2. $(\frac{1}{4} y + 5 x) (\frac{1}{4} y - 5 x)$ $(1/4y+5x)(1/4y-5x)$
3. $(\frac{1}{16} y + 5 x) (\frac{1}{16} y - 5 x)$ $(1/16y+5x)(1/16y-5x)$
4. $(\frac{1}{2} y + 25 x) (\frac{1}{2} y - 25 x)$ $(1/2y+25x)(1/2y-25x)$

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Factoring the Difference of Two Squares

a^{2} - b^{2} = (a + b) (a - b)

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

First, express both terms of the polynomial as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

	$\frac{1}{4} y^{2} - 25 x^{2}$ $1/4y^2-25x^2$

$=$ $=$	${(\frac{1}{2} y)}^{2} - 25 x^{2}$ $(1/2y)^2-25x^2$	${(\frac{1}{2} y)}^{2} = \frac{1}{4} y^{2}$ $(1/2y)^2=1/4y^2$

$=$ $=$	${(\frac{1}{2} y)}^{2} - {(5 x)}^{2}$ $(1/2y)^2-(5x)^2$	${(5 x)}^{2} = 25 x^{2}$ $(5x)^2=25x^2$

Next, label the values in the expression.

a^{2} - b^{2}

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$

{(\frac{1}{2} y)}^{2} - {(5 x)}^{2}

$(1/2y)^2-(5x)^2$

a = \frac{1}{2} y

$a=1/2y$

b = 5 x

$b=5x$

Substitute the values into the formula given for Factoring the Difference of Two Squares.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$ $=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

$\left(\color{#00880A}{\frac{1}{2}y}\right)^2-(\color{#9a00c7}{5x})^2$	$=$ $=$	$\left(\color{#00880A}{\frac{1}{2}y}+\color{#9a00c7}{5x}\right)\left(\color{#00880A}{\frac{1}{2}y}-\color{#9a00c7}{5x}\right)$

(\frac{1}{2} y + 5 x) (\frac{1}{2} y - 5 x)

$(1/2y+5x)(1/2y-5x)$

Question 2 of 4

Factor.

9 n^{2} - \frac{1}{9}

$9n^2-1/9$

1. $(3 n + \frac{1}{4}) (3 n - \frac{1}{5})$ $(3n+1/4)(3n-1/5)$
2. $(3 n + \frac{1}{6}) (3 n - \frac{1}{6})$ $(3n+1/6)(3n-1/6)$
3. $(4 n + \frac{1}{3}) (5 n - \frac{1}{3})$ $(4n+1/3)(5n-1/3)$
4. $(3 n + \frac{1}{3}) (3 n - \frac{1}{3})$ $(3n+1/3)(3n-1/3)$

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Factoring the Difference of Two Squares

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

First, express both terms of the polynomial as perfect squares. In other words, both terms should have

$2$ as their exponent.

	$9n^2-1/9$

$=$	$(3n)^2-1/9$	$(3n)^2=9n^2$

$=$	$(3n)^2-(1/3)^2$	$(1/3)^2=1/9$

Next, label the values in the expression.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$

$(3n)^2-(1/3)^2$

$a=3n$

$b=1/3$

Substitute the values into the formula given for Factoring the Difference of Two Squares.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

$(\color{#00880A}{3n})^2-\left(\color{#9a00c7}{\frac{1}{3}}\right)^2$	$=$	$\left(\color{#00880A}{3n}+\color{#9a00c7}{\frac{1}{3}}\right)\left(\color{#00880A}{3n}-\color{#9a00c7}{\frac{1}{3}}\right)$

$(3n+1/3)(3n-1/3)$

Question 3 of 4

Factor.

$(x^2)/(16)-(y^2)/(36)$

1. $(x/8+y/6)(x/2-y/6)$
2. $(x/4+y/4)(x/6-y/6)$
3. $(x/4+y/6)(x/4-y/6)$
4. $(x/4+y/9)(x/4-y/4)$

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Factoring the Difference of Two Squares

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

First, express both terms of the polynomial as perfect squares. In other words, both terms should have

$2$ as their exponent.

	$(x^2)/(16)-(y^2)/(36)$

$=$	$(x/4)^2-(y^2)/(36)$	$(x/4)^2=(x^2)/(16)$

$=$	$(x/4)^2-(y/6)^2$	$(y/6)^2=(y^2)/(36)$

Next, label the values in the expression.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$

$(x/4)^2-(y/6)^2$

$a=x/4$

$b=y/6$

Substitute the values into the formula given for Factoring the Difference of Two Squares.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

$\left(\color{#00880A}{\frac{x}{4}}\right)^2-\left(\color{#9a00c7}{\frac{y}{6}}\right)^2$	$=$	$\left(\color{#00880A}{\frac{x}{4}}+\color{#9a00c7}{\frac{y}{6}}\right)\left(\color{#00880A}{\frac{x}{4}}-\color{#9a00c7}{\frac{y}{6}}\right)$

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

Question 4 of 4

Factor.

$64x^2-2 1/4$

1. $(1/3x+4)(1/3x-4)$
2. $(4x+1/2)(4x-1/2)$
3. $(8x+3/2)(8x-3/2)$
4. $(8x+3/4)(8x-3/4)$

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Factoring the Difference of Two Squares

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2=(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

First, express both terms of the polynomial as perfect squares. In other words, both terms should have

$2$ as their exponent.

	$64x^2-2 1/4$

$=$	$(8x)^2-2 1/4$	$(8x)^2=64x^2$

$=$	$(8x)^2-9/4$	Convert the second term to an improper fraction

$=$	$(8x)^2-(3/2)^2$	$(3/2)^2=9/4$

Next, label the values in the expression.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$

$(8x)^2-(3/2)^2$

$a=8x$

$b=3/2$

Substitute the values into the formula given for Factoring the Difference of Two Squares.

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$

$(\color{#00880A}{8x})^2-\left(\color{#9a00c7}{\frac{3}{2}}\right)^2$	$=$	$\left(\color{#00880A}{8x}+\color{#9a00c7}{\frac{3}{2}}\right)\left(\color{#00880A}{8x}-\color{#9a00c7}{\frac{3}{2}}\right)$

$(8x+3/2)(8x-3/2)$

Factor Difference of Two Squares 3

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Factoring the Difference of Two Squares

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Factoring the Difference of Two Squares

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Factoring the Difference of Two Squares

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Factoring the Difference of Two Squares

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