Factor Difference of Two Squares (Harder) 2

Question 1 of 4

Factor.

$2x^5-128x^3$

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

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

Start by listing down their factors.

Factors of

$2x^5$ : $2$

$times$ $x$

$timesxtimesx$

Factors of

$128x^3$ : $2$

$times64times$ $x$

$times$ $x$

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

GCF	$=$	$2$ $times$ $x$ $times$ $x$ $times$ $x$
	$=$	$2x^3$

Next, factor by placing

$2x^3$ outside a bracket.

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

$2x^3$ , then simplify.

		$2x^3[(2x^5div2x^3)-(128x^3div2x^3)]$
	$=$	$2x^3(x^2-64)$

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

$2$ as their exponent.

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

Finally, label the values in the expression

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

$a=x$

$b=8$

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

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

Question 2 of 4

Factor.

$3x^2y^2-3$

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

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

Start by listing down their factors.

Factors of

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

$timesxtimesxtimesytimesy$

Factors of

$3$ :

$1times$ $3$

Both

$3x^2y^2$ and

$3$ have

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

Next, factor by placing

$3$ outside a bracket.

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

$3$ , then simplify.

		$3[(3x^2y^2div3)-(3div3)]$
	$=$	$3(x^2y^2-1)$

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

$2$ as their exponent.

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

Finally, label the values in the expression

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

$a=xy$

$b=1$

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

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

Question 3 of 4

Factor.

$m^4-81$

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

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

$2$ as their exponent.

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

Next, label the values in the expression

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

$a=m^2$

$b=9$

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

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

$2$ as their exponent.

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

Finally, label the values in the expression

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

$a=m$

$b=3$

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

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

Question 4 of 4

Factor.

$16x^4-1$

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

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

$2$ as their exponent.

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

Next, label the values in the expression

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

$a=4x^2$

$b=1$

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

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

$2$ as their exponent.

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

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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Correct!

Factoring the Difference of Two Squares

2. Question

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Factoring the Difference of Two Squares

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Exceptional!

Factoring the Difference of Two Squares

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Factoring the Difference of Two Squares

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