Factorise Difference of Two Squares (Harder) 3

Question 1 of 4

Factor.

5 u^{4} - 5

$5u^4-5$

1. $5 (u^{2} + 1) (2 u + 1)$ $5(u^2+1)(2u+1)$
2. $5 (u^{2} + 1) (u + 1)$ $5(u^2+1)(u+1)$
3. $(5 u^{2} + 1) (u - 1) (u + 1)$ $(5u^2+1)(u-1)(u+1)$
4. $5 (u^{2} + 1) (u - 1) (u + 1)$ $5(u^2+1)(u-1)(u+1)$

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Factorising 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

5 u^{4}

$5u^4$ : $5$ $5$

\times u \times u \times u \times u

$timesutimesutimesutimesu$

Factors of

5

$5$ :

1 \times

$1times$ $5$ $5$

Both

5 u^{4}

$5u^4$ and

5

$5$ have

5

$5$ as their factor, so it is the GCF.

Next, factor by placing

5

$5$ outside a bracket.

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

5

$5$ , then simplify.

		$5 [(5 u^{4} \div 5) - (5 \div 5)]$ $5[(5u^4div5)-(5div5)]$
	$=$ $=$	$5 (u^{4} - 1)$ $5(u^4-1)$

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

2

$2$ as their exponent.

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

Next, label the values in the expression

{(u^{2})}^{2} - 1^{2}

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

a = u^{2}

$a=u^2$

b = 2

$b=2$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$ $=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$
$5(\color{#00880A}{u^2})^2-\color{#9a00c7}{1}^2$	$=$ $=$	$5(\color{#00880A}{u^2}+\color{#9a00c7}{1})(\color{#00880A}{u^2}-\color{#9a00c7}{1})$

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

2

$2$ as their exponent.

		$u^{2} - 1$ $u^2-1$
	$=$ $=$	$u^{2} - 1^{2}$ $u^2-1^2$	$1^{2} = 1$ $1^2=1$

Finally, label the values in the expression

u^{2} - 1^{2}

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

a = u

$a=u$

b = 1

$b=1$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$	$(\color{#00880A}{a}-\color{#9a00c7}{b})(\color{#00880A}{a}+\color{#9a00c7}{b})$
$5(u^2+1)(\color{#00880A}{u}^2-\color{#9a00c7}{1}^2)$	$=$	$5(u^2+1)(\color{#00880A}{u}-\color{#9a00c7}{1})(\color{#00880A}{u}+\color{#9a00c7}{1})$

$5(u^2+1)(u-1)(u+1)$

Question 2 of 4

Factor.

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

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

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Factoring the Difference of Two Squares

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

Label the values in the expression

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

$a=x+y$

$b=z$

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

$(x+y+z)(x+y-z)$

Question 3 of 4

Factor.

$y^2-(m-n)^2$

1. $(y+mn)(y-mn)$
2. $(y+m-n)(y-m+n)$
3. $(y+m-n)(y-m-n)$
4. $(y-n)(y+m)$

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Factoring the Difference of Two Squares

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

Label the values in the expression

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

$a=y$

$b=m-n$

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

$(y+m-n)(y-m+n)$

Question 4 of 4

factor.

$4a^2-4(b-c)^2$

1. $4(a-b+c)(b+c)$
2. $4(a+b-c)(a-b+c)$
3. $4(a+b)(a+c)$
4. $4(a-c)(a+b)$

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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

$4a^2$ : $4$

$timesatimesa$

Factors of

$4(b-c)^2$ : $4$

$times(b-c)times(b-c)$

Both

$4a^2$ and

$4(b-c)^2$ have

$4$ as their factor, so it is the GCF.

Next, factor by placing

$4$ outside a bracket.

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

$4$ , then simplify.

		$4[(4a^2div4)-(4(b-c)^2div4)]$
	$=$	$4(a^2-(b-c)^2)$

Since both terms have

$2$ as their exponents, they are perfect squares.

Factor the expression further by using the formula given for Factor the Difference of Two Squares.

$a=a$

$b=(b-c)$

$\color{#00880A}{a}^2-\color{#9a00c7}{b}^2$	$=$	$(\color{#00880A}{a}+\color{#9a00c7}{b})(\color{#00880A}{a}-\color{#9a00c7}{b})$
$4(\color{#00880A}{a}^2-\color{#9a00c7}{(b-c)}^2)$	$=$	$4(\color{#00880A}{a}+\color{#9a00c7}{(b-c)})(\color{#00880A}{a}-\color{#9a00c7}{(b-c)})$
	$=$	$4(a+b-c)(a-b+c)$	Remove the brackets inside

$4(a+b-c)(a-b+c)$

Factorise Difference of Two Squares (Harder) 3

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Quiz summary

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1. Question

Hint

Fantastic!

Factorising the Difference of Two Squares

2. Question

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Correct!

Factoring the Difference of Two Squares

3. Question

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Well Done!

Factoring the Difference of Two Squares

4. Question

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Great Work!

Factoring the Difference of Two Squares

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