Factorise by Grouping in Pairs

Question 1 of 7

Factor.

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

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

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

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Factoring by Grouping in Pairs

a (c + d) + b (c + d) = (a + b) (c + d)

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

Method One

First, label the values in the expression:

a (c + d) + b (c + d)

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

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

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

a = 7

$a=7$

b = x

$b=x$

c = x

$c=x$

d = 5

$d=5$

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)

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

Method Two

First, write the bracketed terms once.

7

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

+ x

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

(x + 5)

$(x+5)$

Then place the coefficients in a separate bracket.

$7$ $7$

(x + 5)

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

(x + 5)

$(x+5)$

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

(x + 5)

$(x+5)$

(7 + x) (x + 5)

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

Question 2 of 7

Factor.

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

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

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

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Factoring by Grouping in Pairs

a (c + d) + b (c + d) = (a + b) (c + d)

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

Method One

First, label the values in the expression:

a (c + d) + b (c + d)

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

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

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

a = 6 a

$a=6a$

b = - n

$b=-n$

c = m

$c=m$

d = 2

$d=2$

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

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

(6 a - n) (m + 2)

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

Method Two

First, write the bracketed terms once.

6 a

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

- n

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

(m + 2)

$(m+2)$

Then place the coefficients in a separate bracket.

$6 a$ $6a$

(m + 2)

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

(m + 2)

$(m+2)$

$(6 a - n)$ $(6a-n)$

(m + 2)

$(m+2)$

(6 a - n) (m + 2)

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

Question 3 of 7

Factor.

a b + 7 a + 3 b + 21

$ab+7a+3b+21$

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

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Factoring by Grouping in Pairs

a (c + d) + b (c + d) = (a + b) (c + d)

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

Method One

First, group the terms into pairs and factor them.

First Pair

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

Second Pair

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

Next, label the values in the expression:

a (c + d) + b (c + d)

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

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

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

a = a

$a=a$

b = 3

$b=3$

c = b

$c=b$

d = 7

$d=7$

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)

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

Method Two

First, group the terms into pairs and factor them.

First Pair

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

Second Pair

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

Next, write the bracketed terms once.

a

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

+ 3

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

(b + 7)

$(b+7)$

Then place the coefficients in a separate bracket.

$a$ $a$

(b + 7)

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

(b + 7)

$(b+7)$

$(a + 3)$ $(a+3)$

(b + 7)

$(b+7)$

(a + 3) (b + 7)

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

Question 4 of 7

Factor.

3 a + 6 b + 8 c a + 16 c b

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

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

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Factoring by Grouping in Pairs

a (c + d) + b (c + d) = (a + b) (c + d)

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

Method One

First, group the terms into pairs and factor them.

First Pair

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

Second Pair

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

Next, label the values in the expression:

a (c + d) + b (c + d)

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

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

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

a = 3

$a=3$

b = 8 c

$b=8c$

c = a

$c=a$

d = 2 b

$d=2b$

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 + 8 c) (a + 2 b)

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

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

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

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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.

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

Factorise by Grouping in Pairs

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Quiz summary

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1. Question

Hint

Keep Going!

Factoring by Grouping in Pairs

2. Question

Hint

Great Work!

Factoring by Grouping in Pairs

3. Question

Hint

Keep Going!

Factoring by Grouping in Pairs

4. Question

Hint

Correct!

Factoring by Grouping in Pairs

5. Question

Hint

Fantastic!

Factoring by Grouping in Pairs

6. Question

Hint

Nice Job!

Factoring by Grouping in Pairs

7. Question

Hint

Well Done!

Factoring by Grouping in Pairs

Quizzes