Factoring — Challenge
Factoring is multiplication run backwards. Always pull out the common factor first — it makes everything after it easier.
What these problems cover
These questions use 7 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Integer — A whole number, positive, negative, or zero.
- Area — Space inside, measured in square units.
- Perfect square — A number that is some whole number squared: 1, 4, 9, 16, 25, …
- Factor — A number that divides evenly into another.
- GCF — Greatest common factor — the largest number dividing both.
- Difference of squares — a2 - b2 = (a - b)(a + b).
- Perfect square trinomial — a2 ± 2ab + b2 = (a ± b)2.
Problem 1
Factorise x2+7x+12.
- (x+2)(x+6)
- (x+1)(x+12)
- (x+3)(x+4)
- (x-3)(x-4)
- (x+5)(x+2)
Worked answer
(x+3)(x+4)
Look for two numbers that multiply to 12 and add to 7.
The pairs multiplying to 12 are 1 and 12, 2 and 6, 3 and 4; only 3 and 4 add to 7.
So the factorisation is (x+3)(x+4).
Problem 2
Factorise x2-5x+6.
- (x+2)(x+3)
- (x-2)(x-3)
- (x-1)(x-6)
- (x-5)(x-1)
- (x-6)(x+1)
Worked answer
(x-2)(x-3)
Two numbers must multiply to +6 and add to -5.
A positive product with a negative sum means both are negative: -2 and -3.
So the factorisation is (x-2)(x-3).
Problem 3
Factorise 2x2+6x completely.
- 2(x2+3x)
- 2x(x+3)
- x(2x+6)
- 2x(x+6)
- 6x(x+1)
Worked answer
2x(x+3)
Both terms share a factor of 2 and a factor of x, so the greatest common factor is 2x.
Dividing each term by 2x gives x and 3, so the answer is 2x(x+3).
2(x2+3x) and x(2x+6) are equal in value but not factorised completely, because each still has a common factor inside the bracket.
Problem 4
Factorise x2-16.
- (x-4)2
- (x+4)2
- (x-8)(x+2)
- (x-4)(x+4)
- (x-16)(x+1)
Worked answer
(x-4)(x+4)
This is a difference of two squares, since 16 = 42 and there is no middle term.
A difference of squares factorises as (x-4)(x+4).
The middle terms -4x and +4x cancel when expanded, which is why there is no x term and why (x-4)2 is wrong.
Problem 5
Factorise x2+2x+1.
- (x-1)2
- (x+1)(x+2)
- (x+1)2
- (x+2)2
- (x+1)(x-1)
Worked answer
(x+1)2
Two numbers must multiply to 1 and add to 2; both are 1.
So the factorisation is (x+1)(x+1) = (x+1)2.
This is a perfect square trinomial, which is why the two factors are identical.
Problem 6
Factorise 3x2-12 completely.
- 3(x2-4)
- 3(x-2)(x+2)
- (3x-6)(x+2)
- 3(x-4)(x+4)
- 3(x-2)2
Worked answer
3(x-2)(x+2)
Take out the common factor 3 first: 3(x2-4).
The bracket is still a difference of squares, so it factorises further into (x-2)(x+2).
The complete factorisation is 3(x-2)(x+2); stopping at 3(x2-4) leaves work undone.
Problem 7
Factorise x2-x-6.
- (x+3)(x-2)
- (x-3)(x+2)
- (x-6)(x+1)
- (x-1)(x+6)
- (x-3)(x-2)
Worked answer
(x-3)(x+2)
Two numbers must multiply to -6 and add to -1.
The pair -3 and +2 works: -3· 2 = -6 and -3+2 = -1.
So the factorisation is (x-3)(x+2). Swapping the signs would give a middle term of +x instead.
Problem 8
Factorise 2x2+5x+3.
- (2x+1)(x+3)
- (2x+3)(x-1)
- (2x+3)(x+1)
- (x+3)(x+1)
- (2x-3)(x-1)
Worked answer
(2x+3)(x+1)
The x2 coefficient is 2, so one bracket starts with 2x and the other with x.
The constants must multiply to 3, so they are 3 and 1 in some order.
Testing (2x+3)(x+1) gives a middle term of 2x+3x = 5x, which matches. The other order gives 6x+x = 7x.