Factor Difference of Two Squares (Harder) 2

Question 1 of 4

Factor.

2 x^{5} - 128 x^{3}

$2x^5-128x^3$

1. $3 x^{3} (x - 2) (x + 6)$ $3x^3(x-2)(x+6)$
2. $2 x^{3} (x - 8) (x + 8)$ $2x^3(x-8)(x+8)$
3. $8 x^{3} (x - 2) (x + 3)$ $8x^3(x-2)(x+3)$
4. $3 x^{2} (x - 8) (x + 8)$ $3x^2(x-8)(x+8)$

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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, find the Greatest Common Factor (GCF) of the two terms.

Start by listing down their factors.

Factors of

2 x^{5}

$2x^5$ : $2$ $2$

\times

$times$ $x$ $x$

\times

$times$ $x$ $x$

\times

$times$ $x$ $x$

\times x \times x

$timesxtimesx$

Factors of

128 x^{3}

$128x^3$ : $2$ $2$

\times 64 \times

$times64times$ $x$ $x$

\times

$times$ $x$ $x$

\times

$times$ $x$ $x$

Collect the common factors and multiply them all to get the GCF.

GCF	$=$ $=$	$2$ $2$ $\times$ $times$ $x$ $x$ $\times$ $times$ $x$ $x$ $\times$ $times$ $x$ $x$
	$=$ $=$	$2 x^{3}$ $2x^3$

Next, factor by placing

2 x^{3}

$2x^3$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by

2 x^{3}

$2x^3$ , then simplify.

		$2 x^{3} [(2 x^{5} \div 2 x^{3}) - (128 x^{3} \div 2 x^{3})]$ $2x^3[(2x^5div2x^3)-(128x^3div2x^3)]$
	$=$ $=$	$2 x^{3} (x^{2} - 64)$ $2x^3(x^2-64)$

Now, express both terms inside the parenthesis as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

		$x^{2} - 64$ $x^2-64$
	$=$ $=$	$x^{2} - 8^{2}$ $x^2-8^2$	$8^{2} = 64$ $8^2=64$

Finally, label the values in the expression

x^{2} - 8^{2}

$x^2-8^2$ and substitute the values into the formula given for Factoring the Difference of Two Squares.

a = x

$a=x$

b = 8

$b=8$

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

2 x^{3} (x - 8) (x + 8)

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

Question 2 of 4

Factor.

3 x^{2} y^{2} - 3

$3x^2y^2-3$

1. $3 (y - x) (y + x)$ $3(y-x)(y+x)$
2. $3 (x y - 1) (x y + 1)$ $3(xy-1)(xy+1)$
3. $1 (x y - 3) (x y + 3)$ $1(xy-3)(xy+3)$
4. $3 (y - 1) (x + 1)$ $3(y-1)(x+1)$

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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, find the Greatest Common Factor (GCF) of the two terms.

Start by listing down their factors.

Factors of

3 x^{2} y^{2}

$3x^2y^2$ : $3$ $3$

\times x \times x \times y \times y

$timesxtimesxtimesytimesy$

Factors of

3

$3$ :

1 \times

$1times$ $3$ $3$

Both

3 x^{2} y^{2}

$3x^2y^2$ and

3

$3$ have

3

$3$ as their factor, so it is the GCF.

Next, factor by placing

3

$3$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by

3

$3$ , then simplify.

		$3 [(3 x^{2} y^{2} \div 3) - (3 \div 3)]$ $3[(3x^2y^2div3)-(3div3)]$
	$=$ $=$	$3 (x^{2} y^{2} - 1)$ $3(x^2y^2-1)$

Now, express both terms inside the parenthesis as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

	$x^{2} y^{2} - 1$ $x^2y^2-1$
$=$ $=$	${(x y)}^{2} - 1$ $(xy)^2-1$	${(x y)}^{2} = x^{2} y^{2}$ $(xy)^2=x^2y^2$
$=$ $=$	${(x y)}^{2} - 1^{2}$ $(xy)^2-1^2$	$1^{2} = 1$ $1^2=1$

Finally, label the values in the expression

{(x y)}^{2} - 1^{2}

$(xy)^2-1^2$ and substitute the values into the formula given for Factoring the Difference of Two Squares.

a = x y

$a=xy$

b = 1

$b=1$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$ $=$	$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$
$3[(\color{#00880A}{xy})^2-\color{#9a00c7}{1}^2]$	$=$ $=$	$3(\color{#00880A}{xy}-\color{#9a00c7}{1})(\color{#00880A}{xy}+\color{#9a00c7}{1})$

3 (x y - 1) (x y + 1)

$3(xy-1)(xy+1)$

Question 3 of 4

Factor.

m^{4} - 81

$m^4-81$

1. $(m^{2} + 9) (m - 3) (m + 3)$ $(m^2+9)(m-3)(m+3)$
2. $(m^{2} + 9) (m^{2} - 3)$ $(m^2+9)(m^2-3)$
3. $(m^{2} + 9) (m - 9) (m + 9)$ $(m^2+9)(m-9)(m+9)$
4. $(m^{3} + 9) (m^{2} - 9)$ $(m^3+9)(m^2-9)$

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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 inside the parenthesis as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

	$m^{4} - 81$ $m^4-81$
$=$ $=$	${(m^{2})}^{2} - 81$ $(m^2)^2-81$	${(m^{2})}^{2} = m^{4}$ $(m^2)^2=m^4$
$=$ $=$	${(m^{2})}^{2} - 9^{2}$ $(m^2)^2-9^2$	$9^{2} = 81$ $9^2=81$

Next, label the values in the expression

{(m^{2})}^{2} - 9^{2}

$(m^2)^2-9^2$ and substitute the values into the formula given for Factoring the Difference of Two Squares.

a = m^{2}

$a=m^2$

b = 9

$b=9$

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

Now, express both terms inside the second parenthesis as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

		$m^{2} - 9$ $m^2-9$
	$=$ $=$	$m^{2} - 3^{2}$ $m^2-3^2$	$3^{2} = 9$ $3^2=9$

Finally, label the values in the expression

m^{2} - 3^{2}

$m^2-3^2$ and substitute the values into the formula given for Factoring the Difference of Two Squares.

a = m

$a=m$

b = 3

$b=3$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$ $=$	$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$
$(m^2+9)(\color{#00880A}{m}^2-\color{#9a00c7}{3}^2)$	$=$ $=$	$(m^2+9)(\color{#00880A}{m}-\color{#9a00c7}{3})(\color{#00880A}{m}+\color{#9a00c7}{3})$

(m^{2} + 9) (m - 3) (m + 3)

$(m^2+9)(m-3)(m+3)$

Question 4 of 4

Factor.

16 x^{4} - 1

$16x^4-1$

1. $(4 x^{2} + 1) (2 x - 1) (2 x + 1)$ $(4x^2+1)(2x-1)(2x+1)$
2. $(4 x^{3} + 1) (2 - 1) (2 x + 1)$ $(4x^3+1)(2-1)(2x+1)$
3. $(4 x^{2} + 1) (4 - 1) (x + 1)$ $(4x^2+1)(4-1)(x+1)$
4. $(2 x^{2} + 1) (2 x - 1) (2 x + 1)$ $(2x^2+1)(2x-1)(2x+1)$

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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 inside the parenthesis as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

	$16 x^{4} - 1$ $16x^4-1$
$=$ $=$	${(4 x^{2})}^{2} - 1$ $(4x^2)^2-1$	${(4 x^{2})}^{2} = 16 x^{4}$ $(4x^2)^2=16x^4$
$=$ $=$	${(4 x^{2})}^{2} - 1^{2}$ $(4x^2)^2-1^2$	$1^{2} = 1$ $1^2=1$

Next, label the values in the expression

{(4 x^{2})}^{2} - 1^{2}

$(4x^2)^2-1^2$ and substitute the values into the formula given for Factoring the Difference of Two Squares.

a = 4 x^{2}

$a=4x^2$

b = 1

$b=1$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$ $=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$
$(\color{#00880A}{4x^2})^2-\color{#9a00c7}{1}^2$	$=$ $=$	$(\color{#00880A}{4x^2}+\color{#9a00c7}{1})(\color{#00880A}{4x^2}-\color{#9a00c7}{1})$

Now, express both terms inside the second parenthesis as perfect squares. In other words, both terms should have

2

$2$ as their exponent.

	$4 x^{2} - 1$ $4x^2-1$
$=$ $=$	${(2 x)}^{2} - 1$ $(2x)^2-1$	${(2 x)}^{2} = 4 x^{2}$ $(2x)^2=4x^2$
$=$ $=$	${(2 x)}^{2} - 1^{2}$ $(2x)^2-1^2$	$1^2=1$

Finally, label the values in the expression

$(2x)^2-1^2$ and substitute the values into the formula given for Factoring the Difference of Two Squares.

$a=2x$

$b=1$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$	$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$
$(4x^2+1)[(\color{#00880A}{2x})^2-\color{#9a00c7}{1}^2]$	$=$	$(4x^2+1)(\color{#00880A}{2x}-\color{#9a00c7}{1})(\color{#00880A}{2x}+\color{#9a00c7}{1})$

$(4x^2+1)(2x-1)(2x+1)$

Factor Difference of Two Squares (Harder) 2

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1. Question

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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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