# Factor by Grouping Introduction,Factor by Grouping Steps

Factor by Grouping Steps
Examples

Factoring by grouping is an approach that can be used to factor some quadratic expressions and equations in Algebra.
What we do is follow appropriate factor by grouping steps which fortunately aren’t too difficult to get the hang of.

When we look to factor quadratic trinomials by grouping, we initially separate the middle term into two terms. Then rewrite the now four terms into pairs, and take out the common factor.

We’ll show how this is done for quadratics of the form   ax^2 + bx + c.

But the handy thing to remember is that we are doing the opposite of expanding brackets with two binomials, when we’re factoring a quadratic trinomial.

(x+2)(x+3)   multiplied out expands to   x^2 + 5x + 6.

(x+2)(x+3)   are the factors of   x^2 + 5x + 6.

With factor by grouping examples we are going the other way.

### Factor by Grouping Steps:

We can observe a standard expression of the form  ax^2 + bx + c  as an example.

x^2 \space {\text{–}} \space x \space {\text{–}} \space 12

The factor by grouping steps we take are as follows.

1)   Multiply  a  and  c  from the outer terms together, noting the result.

1 \times {\text{-}}12 \space = \space {\text{-}}12

2)   Now identify two terms that multiply together to give this  {\text{-}}12.

But also add together to give the middle term, which here is  {\text{-}}1.
Listing the factor pairs of  –12  can help us identify the numbers required.

For  –12  we have,   1  -12  ,  -12  1  ,  -3  4  ,  -4  3.

We can see that  –4  and  3  work.
So we now rewrite the trinomial with the middle term replaced by these numbers.
The order doesn’t matter, we should arrive at the same conclusion.

x^2 \space {\text{–}} \space 4x + 3x \space {\text{–}} \space 12

3)   Now group together the terms of the expression into pairs.

(x^2 \space {\text{–}} \space 4x)(3x \space {\text{–}} \space 12)

Then take a common factor out of each pair.

x(x \space {\text{–}} \space 3)+4(x \space {\text{–}} \space 3)

4)   With both sets of brackets now being the same, the terms can be combined like so,   (x + 4)(x \space {\text{–}} \space 3).

(x + 4)(x \space {\text{–}} \space 3) &nbsp are the factors of   x^2 \space {\text{–}} \space x \space {\text{–}} \space 12

[ It’s not always the case that a quadratic trinomial of this form can be factored, but when it can, this is an effective approach. ]

Examples

(1.1)

Factor by grouping   8x^2 + 10x + 3.

Solution

8 \times 3 \space = \space 24

Factor pairs of  24  are:   1  24  ,  2  12  ,  3  8  ,  4  6.

4  and  6  are the pair that add up to the middle term  10

Now we have:

8x^2 + (4 + 6)x + 3 \space \space = \space \space 8x^2 + 4x + 6x + 3.

(8x^2 + 4x)+(6x + 3) \space \space = \space \space 4x(2x+1) + 3(2x + 1)

Giving us the factors   (4x + 3)(2x + 1).

(1.2)

Factor by grouping   4x^2 + 14x \space {\text{–}}\space 8.

Solution

4 \times {\text{-}}8 \space = \space {\text{-}}32

Factor pairs of  –32&nbsp are:   1  -32  ,  -1  32  ,  2  -16  ,  -16  2  ,  -4  8  ,  8  -4.

16  and  –2  are the pair that add up to the middle term  14

Now we have:

4x^2 + (16 \space {\text{–}} \space 2)x \space {\text{–}} \space 8 \space \space = \space \space 4x^2 + 16x \space {\text{–}} \space 2x \space {\text{–}} \space 8.

(4x^2 + 16x) \space {\text{–}} \space (2x + 8) \space \space = \space \space 4x(x+4) \space {\text{–}} \space 2(x + 4)

Giving us the factors   (4x \space {\text{–}} \space 2)(x + 4).

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