Factor by Grouping

Question 1 of 7

Factor.

$7(x+5)+x(x+5)$

1. $12(x+5)$
2. $(7+x)(x+2)$
3. $(x)(x+12)$
4. $(7+x)(x+5)$

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

Method One

First, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$7(x+5)+x(x+5)$

$a=7$

$b=x$

$c=x$

$d=5$

Substitute the values into the formula given for Factoring by Grouping in Pairs.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$\color{#00880A}{7}(\color{#9a00c7}{x}+\color{#9a00c7}{5})+\color{#00880A}{x}(\color{#9a00c7}{x}+\color{#9a00c7}{5})$	$=$	$(\color{#00880A}{7}+\color{#00880A}{x})(\color{#9a00c7}{x}+\color{#9a00c7}{5})$

$(7+x)(x+5)$

Method Two

First, write the bracketed terms once.

$7$ $(x+5)$

$+x$ $(x+5)$

$(x+5)$

Then place the coefficients in a separate bracket.

$7$

$(x+5)$ $+x$

$(x+5)$

$(7+x)$

$(x+5)$

$(7+x)(x+5)$

Question 2 of 7

Factor.

$6a(m+2)-n(m+2)$

1. $(3a-n)(m+1)$
2. $(6a-2)(m+n)$
3. $(6a+m)(n-2)$
4. $(6a-n)(m+2)$

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

Method One

First, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$6a(m+2)-n(m+2)$

$a=6a$

$b=-n$

$c=m$

$d=2$

Substitute the values into the formula given for Factoring by Grouping in Pairs.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$\color{#00880A}{6a}(\color{#9a00c7}{m}+\color{#9a00c7}{2})\color{#00880A}{-n}(\color{#9a00c7}{m}+\color{#9a00c7}{2})$	$=$	$[\color{#00880A}{6a}+(\color{#00880A}{-n})](\color{#9a00c7}{m}+\color{#9a00c7}{2})$
	$=$	$(6a-n)(m+2)$

$(6a-n)(m+2)$

Method Two

First, write the bracketed terms once.

$6a$ $(m+2)$

$-n$ $(m+2)$

$(m+2)$

Then place the coefficients in a separate bracket.

$6a$

$(m+2)$ $-n$

$(m+2)$

$(6a-n)$

$(m+2)$

$(6a-n)(m+2)$

Question 3 of 7

Factor.

$ab+7a+3b+21$

1. $(a+7)(b+3)$
2. $(b+3)(a+7)$
3. $(a+b)(3+7)$
4. $(a+3)(b+7)$

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

Method One

First, group the terms into pairs and factor them.

First Pair

		$ab+7a\color{#9E9E9E}{+3b+21}$
	$=$	$a(b+7)\color{#9E9E9E}{+3b+21}$	$a(b+7)=ab+7a$

Second Pair

		$\color{#9E9E9E}{a(b+7)}+3b+21$
	$=$	$\color{#9E9E9E}{a(b+7)}+3(b+7)$	$3(b+7)=3b+21$

Next, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$a(b+7)+3(b+7)$

$a=a$

$b=3$

$c=b$

$d=7$

Finally, substitute the values into the formula given for Factoring by Grouping in Pairs.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$\color{#00880A}{a}(\color{#9a00c7}{b}+\color{#9a00c7}{7})+\color{#00880A}{3}(\color{#9a00c7}{b}+\color{#9a00c7}{7})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{3})(\color{#9a00c7}{b}+\color{#9a00c7}{7})$

$(a+3)(b+7)$

Method Two

First, group the terms into pairs and factor them.

First Pair

		$ab+7a\color{#9E9E9E}{+3b+21}$
	$=$	$a(b+7)\color{#9E9E9E}{+3b+21}$	$a(b+7)=ab+7a$

Second Pair

		$\color{#9E9E9E}{a(b+7)}+3b+21$
	$=$	$\color{#9E9E9E}{a(b+7)}+3(b+7)$	$3(b+7)=3b+21$

Next, write the bracketed terms once.

$a$ $(b+7)$

$+3$ $(b+7)$

$(b+7)$

Then place the coefficients in a separate bracket.

$a$

$(b+7)$ $+3$

$(b+7)$

$(a+3)$

$(b+7)$

$(a+3)(b+7)$

Question 4 of 7

Factor.

$3a+6b+8ca+16cb$

1. $(3+8c)(a+2b)$
2. $(3+8c)(b+2a)$
3. $(4+2c)(a+3b)$
4. $(3+8b)(a+2c)$

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

Method One

First, group the terms into pairs and factor them.

First Pair

		$3a+6b\color{#9E9E9E}{+8ca+16cb}$
	$=$	$3(a+2b)\color{#9E9E9E}{+8ca+16cb}$	$3(a+2b)=3a+6b$

Second Pair

		$\color{#9E9E9E}{3(a+2b)}+8ca+16cb$
	$=$	$\color{#9E9E9E}{a(a+2b)}+8c(a+2b)$	$8c(a+2b)=8ca+16cb$

Next, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$3(a+2b)+8c(a+2b)$

$a=3$

$b=8c$

$c=a$

$d=2b$

Finally, substitute the values into the formula given for Factoring by Grouping in Pairs.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$\color{#00880A}{3}(\color{#9a00c7}{a}+\color{#9a00c7}{2b})+\color{#00880A}{8c}(\color{#9a00c7}{a}+\color{#9a00c7}{2b})$	$=$	$(\color{#00880A}{3}+\color{#00880A}{8c})(\color{#9a00c7}{a}+\color{#9a00c7}{2b})$

$(3+8c)(a+2b)$

Method Two

First, group the terms into pairs and factor them.

First Pair

		$3a+6b\color{#9E9E9E}{+8ca+16cb}$
	$=$	$3(a+2b)\color{#9E9E9E}{+8ca+16cb}$	$3(a+2b)=3a+6b$

Second Pair

		$\color{#9E9E9E}{3(a+2b)}+8ca+16cb$
	$=$	$\color{#9E9E9E}{a(a+2b)}+8c(a+2b)$	$8c(a+2b)=8ca+16cb$

Next, write the bracketed terms once.

$3$ $(a+2b)$

$+8c$ $(a+2b)$

$(a+2b)$

Then place the coefficients in a separate bracket.

$3$

$(a+2b)$ $+8c$

$(a+2b)$

$(3+8c)$

$(a+2b)$

$(3+8c)(a+2b)$

Question 5 of 7

Factor.

$x-7y+xz-7yz$

1. $(2x-y)(1+7z)$
2. $(x-7y)(1+z)$
3. $(1-7y)(x+z)$
4. $(7x-y)(2+z)$

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

Method One

First, group the terms into pairs and factor them.

Pair terms with the same variables.

First Pair

		$x\color{#9E9E9E}{-7y}+xz\color{#9E9E9E}{-7yz}$
	$=$	$x(1+z)\color{#9E9E9E}{-7y-7yz}$	$x(1+z)=x+xz$

Second Pair

		$\color{#9E9E9E}{x(1+z)}-7y-7yz$
	$=$	$\color{#9E9E9E}{x(1+z)}-7y(1+z)$	$-7y(1+z)=-7y-7yz$

Next, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$x(1+z)-7y(1+z)$

$a=x$

$b=-7y$

$c=1$

$d=z$

Finally, substitute the values into the formula given for Factoring by Grouping in Pairs.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$\color{#00880A}{x}(\color{#9a00c7}{1}+\color{#9a00c7}{z})-\color{#00880A}{7y}(\color{#9a00c7}{1}+\color{#9a00c7}{z})$	$=$	$[\color{#00880A}{x}+(\color{#00880A}{-7y})](\color{#9a00c7}{1}+\color{#9a00c7}{z})$
	$=$	$(x-7y)(1+z)$

$(x-7y)(1+z)$

Method Two

First, group the terms into pairs and factor them.

Pair terms with the same variables.

First Pair

		$x\color{#9E9E9E}{-7y}+xz\color{#9E9E9E}{-7yz}$
	$=$	$x(1+z)\color{#9E9E9E}{-7y-7yz}$	$x(1+z)=x+xz$

Second Pair

		$\color{#9E9E9E}{x(1+z)}-7y-7yz$
	$=$	$\color{#9E9E9E}{x(1+z)}-7y(1+z)$	$-7y(1+z)=-7y-7yz$

Next, write the bracketed terms once.

$x$ $(1+z)$

$-7y$ $(1+z)$

$(1+z)$

Then place the coefficients in a separate bracket.

$x$

$(1+z)$ $-7y$

$(1+z)$

$(x-7y)$

$(1+z)$

$(x-7y)(1+z)$

Question 6 of 7

Factor.

$m^3+m^2+m+1$

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

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

Method One

First, group the terms into pairs and factor them.

First Pair

		$m^3+m^2\color{#9E9E9E}{+m+1}$
	$=$	$m^2(m+1)\color{#9E9E9E}{+m+1}$	$m^2(m+1)=m^3+m^2$

Second Pair

		$\color{#9E9E9E}{m^2(m+1)}+m+1$
	$=$	$\color{#9E9E9E}{m^2(m+1)}+1(m+1)$	$1(m+1)=m+1$

Next, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$m^2(m+1)+1(m+1)$

$a=m^2$

$b=1$

$c=m$

$d=1$

Finally, substitute the values into the formula given for Factoring by Grouping in Pairs.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$\color{#00880A}{m^2}(\color{#9a00c7}{m}+\color{#9a00c7}{1})+\color{#00880A}{1}(\color{#9a00c7}{m}+\color{#9a00c7}{1})$	$=$	$(\color{#00880A}{m^2}+\color{#00880A}{1})(\color{#9a00c7}{m}+\color{#9a00c7}{1})$

$(m^2+1)(m+1)$

Method Two

First, group the terms into pairs and factor them.

First Pair

		$m^3+m^2\color{#9E9E9E}{+m+1}$
	$=$	$m^2(m+1)\color{#9E9E9E}{+m+1}$	$m^2(m+1)=m^3+m^2$

Second Pair

		$\color{#9E9E9E}{m^2(m+1)}+m+1$
	$=$	$\color{#9E9E9E}{m^2(m+1)}+1(m+1)$	$1(m+1)=m+1$

Next, write the bracketed terms once.

$m^2$ $(m+1)$

$+1$ $(m+1)$

$(m+1)$

Then place the coefficients in a separate bracket.

$m^2$

$(m+1)$ $+1$

$(m+1)$

$(m^2+1)$

$(m+1)$

$(m^2+1)(m+1)$

Question 7 of 7

Factor.

$-4x(y+7)-8x^2(y+7)$

1. $-4x(1+2x)(y+7)$
2. $-8x(2x)(y+7)$
3. $8x(1-x)(y+7)$
4. $4x(2-x)(y+7)$

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Factoring by Grouping in Pairs

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})=(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

A Greatest Common Factor is the factor of two terms with the highest value.

Method One

First, label the values in the expression:

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$

$-4x(y+7)-8x^2(y+7)$

$a=-4x$

$b=-8x^2$

$c=y$

$d=7$

Substitute the values into the formula given for Factoring by Grouping in Pairs, then simplify.

$\color{#00880A}{a}(\color{#9a00c7}{c}+\color{#9a00c7}{d})+\color{#00880A}{b}(\color{#9a00c7}{c}+\color{#9a00c7}{d})$	$=$	$(\color{#00880A}{a}+\color{#00880A}{b})(\color{#9a00c7}{c}+\color{#9a00c7}{d})$
$-4x(y+7)-8x^2(y+7)$	$=$	$\left[\color{#00880A}{-4x}+\left(\color{#00880A}{-8x^2}\right)\right](\color{#9a00c7}{y}+\color{#9a00c7}{7})$
	$=$	$(-4x-8x^2)(y+7)$

Further factor the first bracket,

$(-4x-8x^2)$ , by using the Greatest Common Factor (HCF).

Start by listing down their factors.

Factors of

$-4x$ : $-1$

$times$ $4$

$times$ $x$

Factors of

$-8x^2$ : $-1$

$times2times$ $4$

$times$ $x$

$times x$

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

HCF	$=$	$-1$ $times$ $4$ $times$ $x$
	$=$	$-4x$

Finally, factor by placing

$-4x$ outside a bracket.

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

$-4x$ , then simplify.

		$-4x[(-4xdiv-4x)+(-8x^2div-4x)]$
	$=$	$-4x(1+2x)$

Since this is only the factored first bracket, include the second bracket to get the final answer.

$-4x(1+2x)(y+7)$

Method Two

First, write the bracketed terms once.

$-4x$ $(y+7)$

$-8x^2$ $(y+7)$

$(y+7)$

Then place the coefficients in a separate bracket.

$-4x$

$(y+7)$ $-8x^2$

$(y+7)$

$(-4x-8x^2)$

$(y+7)$

Further factor the first bracket,

$(-4x-8x^2)$ , by using the Greatest Common Factor (HCF).

Start by listing down their factors.

Factors of

$-4x$ : $-1$

$times$ $4$

$times$ $x$

Factors of

$-8x^2$ : $-1$

$times2times$ $4$

$times$ $x$

$times x$

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

HCF	$=$	$-1$ $times$ $4$ $times$ $x$
	$=$	$-4x$

Finally, factor by placing

$-4x$ outside a bracket.

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

$-4x$ , then simplify.

		$-4x[(-4xdiv-4x)+(-8x^2div-4x)]$
	$=$	$-4x(1+2x)$

Since this is only the factored first bracket, include the second bracket to get the final answer.

$-4x(1+2x)(y+7)$

Factor by Grouping

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Factoring by Grouping in Pairs

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Factoring by Grouping in Pairs

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Factoring by Grouping in Pairs

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Factoring by Grouping in Pairs

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