Algebraic division, factor theorem, remainder theorem, factorising cubics.
You can factorise quadratics — that is second nature by now. But what about x3−6x2+11x−6? You cannot use the quadratic formula on a cubic. You cannot spot the factorisation by inspection (at least, not reliably).
The factor theorem gives you a systematic way to crack open any polynomial. Combined with algebraic long division, it lets you factorise cubics completely — a skill you will need throughout A-Level, from sketching curves to integration by partial fractions.
Algebraic long division works exactly like the numerical long division you learnt in primary school — except with polynomials instead of numbers.
To divide f(x) by (x−a):
Divide x3−6x2+11x−6 by (x−2).
Set up the long division. We divide term by term:
Step 1: x3÷x=x2. Multiply: x2(x−2)=x3−2x2. Subtract:
x3−6x2−(x3−2x2)=−4x2
Step 2: Bring down +11x. Now divide −4x2÷x=−4x. Multiply: −4x(x−2)=−4x2+8x. Subtract:
−4x2+11x−(−4x2+8x)=3x
Step 3: Bring down −6. Now divide 3x÷x=3. Multiply: 3(x−2)=3x−6. Subtract:
3x−6−(3x−6)=0
The remainder is 0, so:
x3−6x2+11x−6=(x−2)(x2−4x+3)
We can factorise the quadratic further: x2−4x+3=(x−1)(x−3).
So the complete factorisation is (x−1)(x−2)(x−3).
Divide 2x3+3x2−5x+7 by (x+2).
Step 1: 2x3÷x=2x2. Multiply: 2x2(x+2)=2x3+4x2. Subtract:
2x3+3x2−(2x3+4x2)=−x2
Step 2: Bring down −5x. Divide −x2÷x=−x. Multiply: −x(x+2)=−x2−2x. Subtract:
−x2−5x−(−x2−2x)=−3x
Step 3: Bring down +7. Divide −3x÷x=−3. Multiply: −3(x+2)=−3x−6. Subtract:
−3x+7−(−3x−6)=13
So 2x3+3x2−5x+7=(x+2)(2x2−x−3)+13.
The remainder is 13.
Watch out for missing terms. If you need to divide x3−7x+6 by (x−1), notice there is no x2 term. You must write it as x3+0x2−7x+6 before starting the division. Forgetting this placeholder is one of the most common errors.
Divide x3−7x+6 by (x−1).
There is no x2 term, so we must write a placeholder: x3+0x2−7x+6.
Step 1: x3÷x=x2. Multiply: x2(x−1)=x3−x2. Subtract:
x3+0x2−(x3−x2)=x2
Step 2: Bring down −7x. Divide x2÷x=x. Multiply: x(x−1)=x2−x. Subtract:
x2−7x−(x2−x)=−6x
Step 3: Bring down +6. Divide −6x÷x=−6. Multiply: −6(x−1)=−6x+6. Subtract:
−6x+6−(−6x+6)=0
So x3−7x+6=(x−1)(x2+x−6)=(x−1)(x+3)(x−2).
If you skip the 0x2 placeholder, the columns misalign and every term after it comes out wrong. This is one of the most common errors in polynomial division — always check for missing powers.
Answer at least 3 of 4 correctly to complete this section.
Before we state the theorem — try testing some values of x in the cubic below. For which values does f(a)=0? What do those values have in common with the factors?
Choose a cubic below, then test values to find its factors. The factor theorem says: if f(a)=0, then (x−a) is a factor.
The constant term is −6. Try its factors: x=−3,−2,−1,1,2,3.
The factor theorem is beautifully simple:
If f(a)=0, then (x−a) is a factor of f(x).
And the converse: if (x−a) is a factor of f(x), then f(a)=0.
If (x−a) is a factor of f(x), then f(x)=(x−a)⋅Q(x) for some polynomial Q(x).
Substitute x=a: f(a)=(a−a)⋅Q(a)=0⋅Q(a)=0.
That is the entire proof. The factor theorem is simply saying: if you can divide exactly (remainder zero), then plugging in the root gives zero.
This catches students every year:
The rule is always (x−a). If a is negative, the bracket ends up with a plus sign.
Show that (x−3) is a factor of f(x)=x3−2x2−5x+6.
Evaluate f(3):
f(3)=27−18−15+6=0
Since f(3)=0, by the factor theorem, (x−3) is a factor of f(x). □
Given that (x+1) is a factor of f(x)=2x3+ax2−5x−3, find the value of a.
If (x+1) is a factor, then f(−1)=0:
f(−1)=2(−1)3+a(−1)2−5(−1)−3=0
−2+a+5−3=0
a=0
Answer at least 3 of 4 correctly to complete this section.
The remainder theorem is the general version of the factor theorem:
When f(x) is divided by (x−a), the remainder is f(a).
The factor theorem is the special case where the remainder is zero.
When we divide f(x) by (x−a), we get:
f(x)=(x−a)⋅Q(x)+R
where R is the remainder (a constant, since we are dividing by a linear factor). Substitute x=a:
f(a)=(a−a)⋅Q(a)+R=0+R=R
So the remainder is simply f(a). You can find the remainder without doing any long division at all — just substitute.
Find the remainder when f(x)=x3+3x2−2x+5 is divided by (x−1).
f(1)=1+3−2+5=7
The remainder is 7. No long division needed.
Find the remainder when f(x)=2x3−x2+4x−1 is divided by (x+2).
Note that (x+2)=(x−(−2)), so we evaluate f(−2):
f(−2)=2(−8)−(4)+4(−2)−1=−16−4−8−1=−29
The remainder is −29.
Answer at least 3 of 4 correctly to complete this section.
This is where everything comes together. The strategy for factorising a cubic f(x)=ax3+bx2+cx+d completely:
Factorise f(x)=x3+2x2−5x−6 completely.
Step 1: Find a factor. The constant term is −6, so try ±1,±2,±3,±6.
f(1)=1+2−5−6=−8=0
f(−1)=−1+2+5−6=0✓
So (x+1) is a factor.
Step 2: Divide. Dividing x3+2x2−5x−6 by (x+1):
x3÷x=x2. Multiply: x2(x+1)=x3+x2. Subtract: 2x2−x2=x2.
x2÷x=x. Multiply: x(x+1)=x2+x. Subtract: −5x−x=−6x.
−6x÷x=−6. Multiply: −6(x+1)=−6x−6. Subtract: −6−(−6)=0.
f(x)=(x+1)(x2+x−6)
Step 3: Factorise the quadratic.
x2+x−6=(x+3)(x−2)
Complete factorisation:
f(x)=(x+1)(x+3)(x−2)
Check: (x+1)(x+3)=x2+4x+3. Then (x2+4x+3)(x−2)=x3+4x2+3x−2x2−8x−6=x3+2x2−5x−6 ✓
Factorise f(x)=2x3+3x2−11x−6 completely.
The constant term is −6 and the leading coefficient is 2. Possible rational roots include ±1,±2,±3,±6,±21,±23.
Try integers first:
f(1)=2+3−11−6=−12=0
f(2)=16+12−22−6=0✓
So (x−2) is a factor. Divide 2x3+3x2−11x−6 by (x−2):
2x3÷x=2x2. Multiply: 2x2(x−2)=2x3−4x2. Subtract: 3x2+4x2=7x2.
7x2÷x=7x. Multiply: 7x(x−2)=7x2−14x. Subtract: −11x+14x=3x.
3x÷x=3. Multiply: 3(x−2)=3x−6. Subtract: −6+6=0.
f(x)=(x−2)(2x2+7x+3)
Factorise the quadratic: 2x2+7x+3=(2x+1)(x+3).
Complete factorisation:
f(x)=(x−2)(2x+1)(x+3)
f(x)=x3+ax2+bx−12. Given that (x−2) is a factor of f(x) and the remainder when f(x) is divided by (x+1) is −6, find the values of a and b.
Since (x−2) is a factor, f(2)=0:
8+4a+2b−12=0⟹4a+2b=4⟹2a+b=2...(1)
The remainder when divided by (x+1) is −6, so f(−1)=−6:
−1+a−b−12=−6⟹a−b=7...(2)
From (1): b=2−2a. Substitute into (2):
a−(2−2a)=7⟹3a−2=7⟹a=3
So b=2−6=−4.
Therefore a=3 and b=−4.
f(x)=2x3−7x2−17x+10
(a) Use the factor theorem to show that (x−5) is a factor of f(x).
(b) Factorise f(x) completely.
(c) Hence solve 2x3−7x2−17x+10=0.
(a) f(5)=2(125)−7(25)−17(5)+10=250−175−85+10=0
Since f(5)=0, (x−5) is a factor by the factor theorem. □
(b) Dividing 2x3−7x2−17x+10 by (x−5) gives 2x2+3x−2.
Factorise: 2x2+3x−2=(2x−1)(x+2).
f(x)=(x−5)(2x−1)(x+2)
(c) Setting each factor to zero:
x−5=0⟹x=5
2x−1=0⟹x=21
x+2=0⟹x=−2
The solutions are x=−2,x=21,x=5.
| Theorem | Statement | Use |
|---|---|---|
| Factor theorem | f(a)=0⟺(x−a) is a factor | Finding factors of polynomials |
| Remainder theorem | f(x)÷(x−a) has remainder f(a) | Finding remainders without division |
| Algebraic division | Long division for polynomials | Dividing after finding a factor |
The factorisation strategy: list candidates (factors of constant term) → test with factor theorem → divide → factorise the quadratic.
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Completing the square, discriminant, solving quadratic equations and inequalities.
Decomposing rational expressions into partial fractions with linear and repeated factors.
Binomial expansion for positive integer powers, binomial coefficients, approximations.