Factorise by Grouping in Pairs

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Question 1 of 7

1. Question

Factor.

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

`(7+x)(x+5)`
`(x)(x+12)`
`(7+x)(x+2)`
`12(x+5)`

Hint

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Correct

Keep Going!

Incorrect

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

2. Question

Factor.

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

`(6a-n)(m+2)`
`(6a+m)(n-2)`
`(6a-2)(m+n)`
`(3a-n)(m+1)`

Hint

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Correct

Great Work!

Incorrect

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

3. Question

Factor.

`ab+7a+3b+21`

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

Hint

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Correct

Keep Going!

Incorrect

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

4. Question

Factor.

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

`(3+8c)(a+2b)`
`(4+2c)(a+3b)`
`(3+8b)(a+2c)`
`(3+8c)(b+2a)`

Hint

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Correct

Correct!

Incorrect

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

5. Question

Factor.

`x-7y+xz-7yz`

`(x-7y)(1+z)`
`(1-7y)(x+z)`
`(7x-y)(2+z)`
`(2x-y)(1+7z)`

Hint

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Correct

Fantastic!

Incorrect

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

6. Question

Factor.

`m^3+m^2+m+1`

`(m^2+1)(m+1)`
`(2m+1)(m+1)`
`(2m^2+1)(m+1)`
`(m+1)(2m+1)`

Hint

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Correct

Nice Job!

Incorrect

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

7. Question

Factor.

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

`-4x(1+2x)(y+7)`
`-8x(2x)(y+7)`
`4x(2-x)(y+7)`
`8x(1-x)(y+7)`

Hint

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Correct

Well Done!

Incorrect

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)`

Quizzes

Binomial Products – Distributive Property
Binomial Products – FOIL Method
FOIL Method – Same First Variable 1
FOIL Method – Same First Variable 2
FOIL Method – 3 Terms
Expand Perfect Squares
Difference of Two Squares
Expand Longer Expressions
Highest Common Factor 1
Highest Common Factor 2
Factorise a Polynomial (HCF)
Factorise a Polynomial 1
Factorise a Polynomial 2
Factorise a Polynomial with Integers
Factorise Difference of Two Squares 1
Factorise Difference of Two Squares 2
Factorise Difference of Two Squares 3
Factorise by Grouping in Pairs
Factorise Difference of Two Squares (Harder) 1
Factorise Difference of Two Squares (Harder) 2
Factorise Difference of Two Squares (Harder) 3
Factorise Trinomials (Quadratics) 1
Factorise Trinomials (Quadratics) 2
Factorise Trinomials (Quadratics) 3
Factorise Trinomials (Quadratics) w Coefficient more than 1 (1)
Factorise Trinomials (Quadratics) w Coefficient more than 1 (2)
Factorise Trinomials (Quadratics) w Coefficient more than 1 (3)
Factorise Trinomials (Quadratics) – Complex