Difference of Two Squares (Longer Expressions)

Question 1 of 6

Expand and simplify.

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

1. $2+4x^2$
2. $-4x+8$
3. $4x-8$
4. $2x^2-4$

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Expanding Perfect Square Binomials

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

Difference of Two Squares

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

First, expand the values inside the first bracket by Expanding Perfect Square Binomials

$(\color{#00880A}{x}-\color{#9a00c7}{2})^2 \color{#9E9E9E}{-(x-2)(x+2)}$	$=$	$\color{#00880A}{x}^2-2(\color{#00880A}{x})(\color{#9a00c7}{2})+\color{#9a00c7}{2}^2 \color{#9E9E9E}{-(x-2)(x+2)}$
	$=$	$x^2-4x+4-(x-2)(x+2)$	Simplify

Next, expand the values inside the remaining brackets using the Difference of Two Squares

$\color{#9E9E9E}{x^2-4x+4\;-\;}(\color{#00880A}{x}-\color{#9a00c7}{2})(\color{#00880A}{x}+\color{#9a00c7}{2})$	$=$	$\color{#9E9E9E}{x^2-4x+4\;-\;(}\color{#00880A}{x}^2-\color{#9a00c7}{2}^2\color{#9E9E9E}{)}$
	$=$	$x^2-4x+4-(x^2-4)$	Simplify
	$=$	$x^2-4x+4-x^2+4$

Finally, simplify further by combining like terms.

	$x^2$ $-4x+4$ $-x^2$ $+4$
$=$	$-4x$ $+4+4$	$x^2-x^2$ cancels out
$=$	$-4x+8$

$-4x+8$

Question 2 of 6

Expand and simplify.

$28x-(x-2)(x-5)+x^2$

1. $35x-10$
2. $2x^2+35x-10$
3. $2x^2+21x-10$
4. $x^2+17x+15$

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First, use the FOIL Method to expand and simplify the values inside the brackets.

Multiply the First, Outside, Inside, and Last terms, then simplify.

	$\color{#9E9E9E}{28x-[}\color{#00880A}{(x)(x)}+\color{#9a00c7}{(x)(-5)}+\color{#007DDC}{(-2)(x)}+\color{#e65021}{(-2)(-5)}\color{#9E9E9E}{]+x^2}$
$=$	$\color{#9E9E9E}{28x-(}x^2-5x-2x+10\color{#9E9E9E}{)+x^2}$
$=$	$28x-(x^2-7x+10)+x^2$	Collect like terms
$=$	$28x-x^2+7x-10+x^2$

Finally, simplify further by combining like terms.

	$28x$ $-x^2$ $+7x-10$ $+x^2$
$=$	$28x+7x$ $-10$	$x^2-x^2$ cancels out
$=$	$35x-10$

$35x-10$

Question 3 of 6

Expand and simplify.

$(b-a)(b-t)-b^2+bt$

1. $2at-b$
2. $2b^2+ab$
3. $b^2-ab$
4. $at-ab$

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First, use the FOIL Method to expand and simplify the values inside the brackets.

Multiply the First, Outside, Inside, and Last terms, then simplify.

		$\color{#00880A}{(b)(b)}+\color{#9a00c7}{(b)(-t)}+\color{#007DDC}{(-a)(b)}+\color{#e65021}{(-a)(-t)}\color{#9E9E9E}{-b^2+bt}$
	$=$	$b^2-bt-ab+at-b^2+bt$

Finally, simplify further by combining like terms.

	$b^2$ $-bt-ab+at$ $-b^2$ $+bt$
$=$	$-bt$ $-ab+at$ $+bt$	$b^2-b^2$ cancels out
$=$	$-ab+at$	$-bt+bt$ cancels out
$=$	$at-ab$	Rearrange the values

$at-ab$

Question 4 of 6

Expand and simplify.

$(2x+3)(6x-2)-5(x-4)(x+4)$

1. $7x^2-22x+74$
2. $14x^2+7x+86$
3. $7x^2+14x+74$
4. $12x^2-18x-86$

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Difference of Two Squares

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

First, use the FOIL Method to expand and simplify the values inside the first two brackets.

Multiply the First, Outside, Inside, and Last terms, then simplify.

	$\color{#00880A}{(2x)(6x)}+\color{#9a00c7}{(2x)(-2)}+\color{#007DDC}{(3)(6x)}+\color{#e65021}{(3)(-2)}\color{#9E9E9E}{-5(x-4)(x+4)}$
$=$	$12x^2\color{#CC0000}{-4x+18x}-6\color{#9E9E9E}{-5(x-4)(x+4)}$	Collect like terms
$=$	$12x^2+14x-6-5(x-4)(x+4)$

Next, substitute the values inside the remaining brackets into the formula given for Difference of Two Squares.

$\color{#9E9E9E}{12x^2+14x-6-5}(\color{#00880A}{x}-\color{#9a00c7}{4})(\color{#00880A}{x}+\color{#9a00c7}{4})$	$=$	$\color{#9E9E9E}{12x^2+14x-6-5(}\color{#00880A}{x}^2-\color{#9a00c7}{4}^2\color{#9E9E9E}{)}$
	$=$	$12x^2+14x-6-5(x^2-16)$

Then, distribute $5$ to each value inside the bracket.

$12^2+14x-6-$ $5$ $(x^2-16)$	$=$	$12^2+14x-6-($ $5$ $timesx^2)-($ $5$ $times16)$
	$=$	$12^2+14x-6-(5x^2-80)$
	$=$	$12^2+14x-6-5x^2+80$

Finally, simplify further by combining like terms.

		$12x^2$ $+14x-6$ $-5x^2$ $+80$
	$=$	$7x^2+14x$ $-6+80$
	$=$	$7x^2+14x+74$

$7x^2+14x+74$

Question 5 of 6

Expand and simplify.

$-10m-(m-5)^2$

1. $2m^2-20m+25$
2. $-m^2-25$
3. $2m^2+25$
4. $-m^2+20m-25$

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Expanding Perfect Square Binomials

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

Difference of Two Squares

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

First, expand the values inside the bracket by Expanding Perfect Square Binomials

$\color{#9E9E9E}{-\;10m\;-\;}(\color{#00880A}{m}-\color{#9a00c7}{5})^2$	$=$	$\color{#9E9E9E}{-\;10m\;-\;}\color{#00880A}{m}^2-2(\color{#00880A}{m})(\color{#9a00c7}{5})+\color{#9a00c7}{5}^2$
	$=$	$-10m-(m^2-10m+25)$	Simplify
	$=$	$-10m-m^2+10m-25$

Finally, simplify further by combining like terms.

		$-10m$ $-m^2$ $+10m$ $-25$
	$=$	$-m^2-25$	$-10m+10m$ cancels out

$-m^2-25$

Question 6 of 6

Expand and simplify.

$(y-2)^2+7y(y-2)(y+2)$

1. $y^3-7y^2+16y+8$
2. $3y^3+6y^2-32y-8$
3. $y^3+y^2+24y$
4. $7y^3+y^2-32y+4$

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Expanding Perfect Square Binomials

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

Difference of Two Squares

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

First, expand the values inside the first bracket by Expanding Perfect Square Binomials

$(\color{#00880A}{y}-\color{#9a00c7}{2})^2 \color{#9E9E9E}{+7y(y-2)(y+2)}$	$=$	$\color{#00880A}{y}^2-2(\color{#00880A}{y})(\color{#9a00c7}{2})+\color{#9a00c7}{2}^2 \color{#9E9E9E}{+7y(y-2)(y+2)}$
	$=$	$y^2-4y+4+7y(y-2)(y+2)$	Simplify

Next, expand the values inside the remaining brackets using the Difference of Two Squares

$\color{#9E9E9E}{y^2-4y+4+7y}(\color{#00880A}{y}-\color{#9a00c7}{2})(\color{#00880A}{y}+\color{#9a00c7}{2})$	$=$	$\color{#9E9E9E}{y^2-4y+4+7y(}\color{#00880A}{y}^2-\color{#9a00c7}{2}^2\color{#9E9E9E}{)}$
	$=$	$y^2-4y+4+7y(y^2-4)$	Simplify

Then, distribute $7y$ to each value inside the bracket.

$y^2-4y+4+$ $7y$ $(y^2-4)$	$=$	$y^2-4y+4+($ $7y$ $timesy^2)-($ $7y$ $times4)$
	$=$	$y^2-4y+4+7y^3-28y$

Finally, simplify further by combining like terms.

	$y^2$ $-4y$ $+4+7y^3$ $-28y$
$=$	$y^2-32y+4+7y^3$
$=$	$7y^3+y^2-32y+4$	Rearrange the values

$7y^3+y^2-32y+4$

Difference of Two Squares (Longer Expressions)

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1. Question

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Expanding Perfect Square Binomials

Difference of Two Squares

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4. Question

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Difference of Two Squares

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Expanding Perfect Square Binomials

Difference of Two Squares

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Expanding Perfect Square Binomials

Difference of Two Squares

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