Worked examples
The question
Question: factor completely 2x² − 8.
Completely means: until no part can be factored any more.
2x2 − 8
1 Common factor first
All the numbers can be divided by 2.
So the common factor is 2.
2, 8: all in the 2 times table
We divide every term by it.
What is left goes in the bracket.
2x2 ÷ 2 = x2
−8 ÷ 2 = −4
So it is 2(x² − 4).
= 2(x2 − 4)
2 Count the terms
The bracket has 2 terms.
With 2 terms we look for two squares or two cubes.
x² − 4 is one square take away another square.
x² is x squared.
4 is 2 squared.
A square minus a square is the plus bracket times the minus bracket.
a² − b² = (a + b)(a − b)
x2 = (x)², 4 = 2²
So we get 2(x + 2)(x − 2).
= 2(x + 2)(x − 2)
3 Check each bracket
Brackets like x + 2 have just x and a number.
They can't be factored.
x + 2, x − 2: just x and a number
No bracket can be factored any more.
So we are done.
Careful!
Don't forget the common factor 2 in the answer.
Careful: not (x² − 4) ✗
4 Check
Let's check: we multiply it back out.
Each part of one bracket times each part of the other.
Check: (x + 2)(x − 2) = x2 − 2x + 2x − 4
= x2 − 4
Then the common factor times every part.
2(x2 − 4) = 2x2 − 8
We get 2x² − 8 again.
So it is right!
Let's look back.
First: common factor first.
Then: count the terms.
Then: check each bracket.
Last: check.
Answer x + 2, x − 2: just x and a number
▶ Watch it solved
The question
Question: factor completely 2x³ + 10x² + 12x.
Completely means: until no part can be factored any more.
2x3 + 10x2 + 12x
1 Common factor first
All the numbers can be divided by 2.
Every term has x.
So the common factor is 2x.
2, 10, 12: all in the 2 times table
every term has x
We divide every term by it.
What is left goes in the bracket.
2x3 ÷ 2x = x2
10x2 ÷ 2x = 5x
12x ÷ 2x = 6
So it is 2x(x² + 5x + 6).
= 2x(x2 + 5x + 6)
2 Count the terms
The bracket has 3 terms.
With 3 terms we look for two numbers that multiply to c and add to b.
We need two numbers.
They multiply to 6, and they add to 5.
? × ? = 6
? + ? = 5
They multiply to a plus number, so the two numbers have the same sign.
They add to a plus number, so both are plus.
× is + → same signs: both +
We list the pairs that make 6, smallest first.
We add each pair.
We need 5.
2 and 3 give 5.
Yes!
1 × 6: 1 + 6 = 7
2 × 3: 2 + 3 = 5 ✓
So we get 2x(x + 2)(x + 3).
= 2x(x + 2)(x + 3)
3 Check each bracket
Brackets like x + 2 have just x and a number.
They can't be factored.
x + 2, x + 3: just x and a number
No bracket can be factored any more.
So we are done.
Careful!
Don't forget the common factor 2x in the answer.
Careful: not (x² + 5x + 6) ✗
4 Check
Let's check: we multiply it back out.
Each part of one bracket times each part of the other.
Check: (x + 2)(x + 3) = x2 + 3x + 2x + 6
= x2 + 5x + 6
Then the common factor times every part.
2x(x2 + 5x + 6) = 2x3 + 10x2 + 12x
We get 2x³ + 10x² + 12x again.
So it is right!
Let's look back.
First: common factor first.
Then: count the terms.
Then: check each bracket.
Last: check.
Answer x + 2, x + 3: just x and a number
▶ Watch it solved
The question
Question: factor completely x⁴ − 81.
Completely means: until no part can be factored any more.
x4 − 81
1 Common factor first
Always ask this first.
Here no number and no x goes into every term.
So there is no common factor.
2 Count the terms
The bracket has 2 terms.
With 2 terms we look for two squares or two cubes.
x⁴ − 81 is one square take away another square.
x⁴ is x² squared.
81 is 9 squared.
A square minus a square is the plus bracket times the minus bracket.
a² − b² = (a + b)(a − b)
x4 = (x2)², 81 = 9²
So we get (x² + 9)(x² − 9).
= (x2 + 9)(x2 − 9)
3 Factor again
x² − 9 is one square take away another square.
x² is x squared.
9 is 3 squared.
A square minus a square is the plus bracket times the minus bracket.
a² − b² = (a + b)(a − b)
x2 = (x)², 9 = 3²
So we get (x² + 9)(x + 3)(x − 3).
= (x2 + 9)(x + 3)(x − 3)
4 Check each bracket
Brackets like x + 3 have just x and a number.
They can't be factored.
x + 3, x − 3: just x and a number
x² + 9 is a square PLUS a square.
A sum of two squares does not factor.
We call it prime.
a² + b² → prime
No bracket can be factored any more.
So we are done.
Careful!
That is not the end: x² − 9 is two squares again.
Careful: not (x² + 9)(x² − 9) ✗
5 Check
Let's check: we multiply it back out.
Each part of one bracket times each part of the other.
Check: (x + 3)(x − 3) = x2 − 3x + 3x − 9
= x2 − 9
Then the next bracket.
(x2 + 9)(x2 − 9) = x4 − 9x2 + 9x2 − 81
= x4 − 81
We get x⁴ − 81 again.
So it is right!
Let's look back.
First: common factor first.
Then: count the terms.
Then: factor again.
Then: check each bracket.
Last: check.
▶ Watch it solved
