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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Correct!

Factoring the Difference of Two Squares

2. Question

Hint

Great Work!

Factoring the Difference of Two Squares

3. Question

Hint

Exceptional!

Factoring the Difference of Two Squares

4. Question

Hint

Nice Job!

Factoring the Difference of Two Squares

Quizzes