Worked examples
The question
Question: factor x³ + 2x² + 3x + 6.
There are four terms.
So we factor them in pairs.
This is called grouping.
x3 + 2x2 + 3x + 6
1 Group in pairs
We put the first two together, and the last two together.
= (x3 + 2x2) + (3x + 6)
From the first pair we take out x².
From the second pair we take out 3.
Both leave the same bracket, x + 2.
(x3 + 2x2) = x2(x + 2)
(3x + 6) = 3(x + 2)
Now both parts have x + 2 in them.
= x2(x + 2) + 3(x + 2)
Careful!
This is not the answer yet.
It is still two parts added.
Factored means one thing times another.
Careful: not done yet: still a + between two parts
2 Same bracket out
x + 2 is in both parts, so we take it out once.
What is left, x² + 3, goes in the other bracket.
ab + cb = b(a + c)
So we get (x + 2)(x² + 3).
= (x + 2)(x2 + 3)
3 Check
Let's check: we multiply it back out.
Each part of the first bracket times each part of the second.
Check: (x + 2)(x2 + 3) = x3 + 3x + 2x2 + 6
= x3 + 2x2 + 3x + 6
We get x³ + 2x² + 3x + 6 again.
So the answer is right!
Let's look back.
First: group in pairs.
Then: same bracket out.
Last: check.
▶ Watch it solved
The question
Question: factor x³ + 4x² − 3x − 12.
There are four terms.
So we factor them in pairs.
This is called grouping.
x3 + 4x2 − 3x − 12
1 Group in pairs
We put the first two together, and the last two together.
= (x3 + 4x2) + (−3x − 12)
From the first pair we take out x².
The second pair starts with a minus.
So we take out −3, with the minus.
Both leave the same bracket, x + 4.
(x3 + 4x2) = x2(x + 4)
(−3x − 12) = −3(x + 4)
Careful!
If we take out 3, without the minus, we get −x − 4 in the bracket.
That does not match x + 4.
Take the minus out too.
Careful: not 3(−x − 4) ✗
Now both parts have x + 4 in them.
= x2(x + 4) − 3(x + 4)
2 Same bracket out
x + 4 is in both parts, so we take it out once.
What is left, x² − 3, goes in the other bracket.
ab + cb = b(a + c)
So we get (x + 4)(x² − 3).
= (x + 4)(x2 − 3)
3 Check
Let's check: we multiply it back out.
Each part of the first bracket times each part of the second.
Check: (x + 4)(x2 − 3) = x3 − 3x + 4x2 − 12
= x3 + 4x2 − 3x − 12
We get x³ + 4x² − 3x − 12 again.
So the answer is right!
Let's look back.
First: group in pairs.
Then: same bracket out.
Last: check.
▶ Watch it solved
The question
Question: factor 2x³ + 6x² + 4x + 12.
There are four terms.
So we factor them in pairs.
This is called grouping.
2x3 + 6x2 + 4x + 12
1 Take out the common factor
All four terms can be divided by 2.
So we take 2 out first, like in the common factor lesson.
2, 6, 4, 12: all in the 2 times table
So it is 2(x³ + 3x² + 2x + 6).
Now we group what is in the bracket.
= 2(x3 + 3x2 + 2x + 6)
2 Group in pairs
We put the first two together, and the last two together.
= 2[(x3 + 3x2) + (2x + 6)]
From the first pair we take out x².
From the second pair we take out 2.
Both leave the same bracket, x + 3.
(x3 + 3x2) = x2(x + 3)
(2x + 6) = 2(x + 3)
Now both parts have x + 3 in them.
= 2[x2(x + 3) + 2(x + 3)]
Careful!
This is not the answer yet.
It is still two parts added.
Factored means one thing times another.
Careful: not done yet: still a + between two parts
3 Same bracket out
x + 3 is in both parts, so we take it out once.
What is left, x² + 2, goes in the other bracket.
ab + cb = b(a + c)
So we get 2(x + 3)(x² + 2).
= 2(x + 3)(x2 + 2)
4 Check
Let's check: we multiply it back out.
Each part of the first bracket times each part of the second.
Then 2 times every term.
Check: 2(x + 3)(x2 + 2)
= 2(x3 + 2x + 3x2 + 6) = 2(x3 + 3x2 + 2x + 6)
= 2x3 + 6x2 + 4x + 12
We get 2x³ + 6x² + 4x + 12 again.
So the answer is right!
Let's look back.
First: take out the common factor.
Then: group in pairs.
Then: same bracket out.
Last: check.
Answer = 2(x + 3)(x2 + 2)
▶ Watch it solved
