Factor by Grouping

Question 1 of 7

Factor.

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

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

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

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

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

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

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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

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

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, write the bracketed terms once.

3

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

+ 8 c

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

(a + 2 b)

$(a+2b)$

Then place the coefficients in a separate bracket.

$3$ $3$

(a + 2 b)

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

(a + 2 b)

$(a+2b)$

$(3 + 8 c)$ $(3+8c)$

(a + 2 b)

$(a+2b)$

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

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

Question 5 of 7

Factor.

x - 7 y + x z - 7 y z

$x-7y+xz-7yz$

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

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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.

Pair terms with the same variables.

First Pair

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

Second Pair

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

Next, label the values in the expression:

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

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

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

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

a = x

$a=x$

b = - 7 y

$b=-7y$

c = 1

$c=1$

d = z

$d=z$

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

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

(x - 7 y) (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

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

Second Pair

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

Next, write the bracketed terms once.

x

$x$ $(1 + z)$ $(1+z)$

- 7 y

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

(1 + z)

$(1+z)$

Then place the coefficients in a separate bracket.

$x$ $x$

(1 + z)

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

(1 + z)

$(1+z)$

$(x - 7 y)$ $(x-7y)$

(1 + z)

$(1+z)$

(x - 7 y) (1 + z)

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

Question 6 of 7

Factor.

m^{3} + m^{2} + m + 1

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

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

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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, group the terms into pairs and factor them.

First Pair

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

Second Pair

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

Next, label the values in the expression:

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

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

m^{2} (m + 1) + 1 (m + 1)

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

a = m^{2}

$a=m^2$

b = 1

$b=1$

c = m

$c=m$

d = 1

$d=1$

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)

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

Method Two

First, group the terms into pairs and factor them.

First Pair

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

Second Pair

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

Next, write the bracketed terms once.

m^{2}

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

+ 1

$+1$ $(m + 1)$ $(m+1)$

(m + 1)

$(m+1)$

Then place the coefficients in a separate bracket.

$m^{2}$ $m^2$

(m + 1)

$(m+1)$ $+ 1$ $+1$

(m + 1)

$(m+1)$

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

(m + 1)

$(m+1)$

(m^{2} + 1) (m + 1)

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

Question 7 of 7

Factor.

- 4 x (y + 7) - 8 x^{2} (y + 7)

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

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

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Factoring by Grouping in Pairs

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

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

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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Quiz summary

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1. Question

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Keep Going!

Factoring by Grouping in Pairs

2. Question

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Great Work!

Factoring by Grouping in Pairs

3. Question

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Keep Going!

Factoring by Grouping in Pairs

4. Question

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Correct!

Factoring by Grouping in Pairs

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Fantastic!

Factoring by Grouping in Pairs

6. Question

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Nice Job!

Factoring by Grouping in Pairs

7. Question

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Well Done!

Factoring by Grouping in Pairs

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