🧮 Polynomials and the Factor Theorem

A Level Maths Year 1 · Algebra and Functions

Level 3 · Ages 16–17

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1 What a Polynomial Is

DfE content area B (Algebra and functions) asks you to manipulate polynomials algebraically: expanding brackets, collecting like terms, factorising, simple algebraic division, and using the factor theorem. These are the tools that let you solve cubic equations and sketch cubic graphs.

A polynomial in $x$ is a finite sum of terms of the form $kx^n$, where each index $n$ is a non-negative whole number ($0, 1, 2, 3, \ldots$). The degree is the highest power that appears with a non-zero coefficient.
PolynomialDegreeNameLeading coefficient
$5 - 2x$$1$Linear$-2$
$3x^2 - x + 7$$2$Quadratic$3$
$x^3 - 6x^2 + 4$$3$Cubic$1$
$2x^4 - x$$4$Quartic$2$
Not polynomials: $\dfrac{3}{x} = 3x^{-1}$ (negative index), $\sqrt{x} = x^{1/2}$ (fractional index), $2^x$ (the variable is in the index). You meet these in Indices and Surds and later in exponentials.
Function notation. Polynomials are usually named $\mathrm{f}(x)$, $\mathrm{p}(x)$ and so on. $\mathrm{f}(2)$ simply means "substitute $x = 2$". This notation is exactly what the factor theorem is written in.
2 Expanding Brackets and Collecting Like Terms

To multiply two polynomials, multiply every term of the first by every term of the second, then collect terms with the same power of $x$. With three brackets, multiply two together first, then multiply the result by the third.

Two quick checks on any expansion
degree of product $=$ sum of the degrees  ·  constant term $=$ product of the constants
Worked Example 1 — A linear times a quadratic

Expand and simplify $(2x - 1)\left(x^2 + 3x - 4\right)$.

①$2x\left(x^2 + 3x - 4\right) = 2x^3 + 6x^2 - 8x$
②$-1\left(x^2 + 3x - 4\right) = -x^2 - 3x + 4$
③Collect: $2x^3 + (6 - 1)x^2 + (-8 - 3)x + 4$
④$= 2x^3 + 5x^2 - 11x + 4$
Check: degree $1 + 2 = 3$ ✓; constant $(-1)(-4) = 4$ ✓.
Worked Example 2 — Three brackets

Expand $(x + 1)(x - 2)(2x + 3)$.

①First pair: $(x+1)(x-2) = x^2 - x - 2$
②Multiply by $(2x + 3)$: $2x\left(x^2 - x - 2\right) + 3\left(x^2 - x - 2\right)$
③$= 2x^3 - 2x^2 - 4x + 3x^2 - 3x - 6$
④$= 2x^3 + x^2 - 7x - 6$
Check: leading coefficient $1 \times 1 \times 2 = 2$ ✓; constant $1 \times (-2) \times 3 = -6$ ✓.
Subtracting a polynomial changes every sign. $\left(x^3 + 2x - 5\right) - \left(x^3 - 4x + 1\right) = 6x - 6$, not $-2x - 4$. Put the second polynomial in brackets before you subtract.
3 Factorising Polynomials

Factorising is expanding in reverse. For a cubic, try these in order:

Worked Example 3 — Common factor, then grouping

Factorise fully (a) $2x^3 - 8x$   (b) $x^3 + 2x^2 - 9x - 18$.

①(a) Common factor $2x$: $2x\left(x^2 - 4\right)$.
②Difference of two squares: $2x(x - 2)(x + 2)$.
③(b) Group in pairs: $x^2(x + 2) - 9(x + 2)$.
④Common bracket: $(x + 2)\left(x^2 - 9\right) = (x + 2)(x - 3)(x + 3)$.
"Factorise fully" means keep going. Stopping at $2x\left(x^2-4\right)$ loses a mark.
4 Simple Algebraic Division

At this stage you only need to divide a polynomial by a linear expression $(x \pm p)$, usually when it is known to be a factor. Dividing a cubic by a linear factor leaves a quadratic.

Method A — long division
Divide the leading terms → multiply back → subtract → bring down → repeat
Worked Example 4 — Long division

Divide $2x^3 - 3x^2 - 11x + 6$ by $(x - 3)$, and hence factorise it fully.

①$2x^3 \div x = 2x^2$.  $2x^2(x - 3) = 2x^3 - 6x^2$.  Subtract: $3x^2 - 11x$.
②$3x^2 \div x = 3x$.  $3x(x - 3) = 3x^2 - 9x$.  Subtract: $-2x + 6$.
③$-2x \div x = -2$.  $-2(x - 3) = -2x + 6$.  Subtract: $0$.
④Quotient $2x^2 + 3x - 2$, remainder $0$.
⑤$2x^2 + 3x - 2 = (2x - 1)(x + 2)$, so $2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x - 1)(x + 2)$.
Method B — comparing coefficients
Write $\mathrm{f}(x) \equiv (x - a)\left(Ax^2 + Bx + C\right)$, expand, and match each power of $x$
Worked Example 5 — A missing term

Given that $(x + 1)$ is a factor of $x^3 - 7x - 6$, find the quadratic factor by comparing coefficients.

①Write $x^3 + 0x^2 - 7x - 6 \equiv (x + 1)\left(x^2 + Bx + C\right)$. The leading coefficient must be $1$.
②Right side: $x^3 + (B + 1)x^2 + (C + B)x + C$.
③Constant: $C = -6$.  $x^2$: $B + 1 = 0$, so $B = -1$.
④Check $x$: $C + B = -6 - 1 = -7$ ✓
⑤$x^3 - 7x - 6 = (x + 1)\left(x^2 - x - 6\right) = (x + 1)(x - 3)(x + 2)$.
In long division, write the missing $x^2$ term as $0x^2$ to keep the columns lined up. Forgetting it is the most common division error.
5 The Factor Theorem
The factor theorem
$\mathrm{f}(a) = 0 \iff (x - a)$ is a factor of $\mathrm{f}(x)$
More generally: $\mathrm{f}\!\left(\tfrac{b}{a}\right) = 0 \iff (ax - b)$ is a factor
Why it is true. Dividing $\mathrm{f}(x)$ by $(x - a)$ always gives $\mathrm{f}(x) \equiv (x - a)\,\mathrm{q}(x) + r$ for some quotient $\mathrm{q}(x)$ and constant remainder $r$. Substituting $x = a$ kills the first term, so $\mathrm{f}(a) = r$. The remainder is zero — that is, $(x - a)$ is a factor — exactly when $\mathrm{f}(a) = 0$.
Mind the sign. To test $(x + 3)$, substitute $x = -3$. To test $(2x - 1)$, substitute $x = \tfrac12$ — the value that makes the bracket zero.
Which values to try? If the leading coefficient is $1$, any integer root must be a factor of the constant term. So for $x^3 + 2x^2 - 5x - 6$ try $\pm1, \pm2, \pm3, \pm6$. A calculator's table function speeds this up — but write down the substitution that gives $0$.
x y −3 −1 2 −6 y = (x+1)(x+3)(x−2) (0, −6) Each root of f(x) = 0 gives a factor: x = −3, −1, 2 ↔ (x+3), (x+1), (x−2)
Worked Example 6 — Factorise and solve a cubic

$\mathrm{f}(x) = x^3 + 2x^2 - 5x - 6$. Factorise $\mathrm{f}(x)$ fully and solve $\mathrm{f}(x) = 0$.

①$\mathrm{f}(1) = 1 + 2 - 5 - 6 = -8 \neq 0$.  $\mathrm{f}(-1) = -1 + 2 + 5 - 6 = 0$ ✓
②By the factor theorem $(x + 1)$ is a factor.
③Divide: $\mathrm{f}(x) = (x + 1)\left(x^2 + x - 6\right)$
④$x^2 + x - 6 = (x + 3)(x - 2)$, so $\mathrm{f}(x) = (x + 1)(x + 3)(x - 2)$.
⑤Solutions: $x = -3,\; -1,\; 2$ — the three crossing points on the graph above.
Worked Example 7 — A factor of the form $(ax - b)$

Show that $(2x - 1)$ is a factor of $\mathrm{f}(x) = 2x^3 + x^2 - 13x + 6$, and factorise $\mathrm{f}(x)$ fully.

①$2x - 1 = 0$ when $x = \tfrac12$.
②$\mathrm{f}\!\left(\tfrac12\right) = 2\left(\tfrac18\right) + \tfrac14 - \tfrac{13}{2} + 6 = \tfrac14 + \tfrac14 - \tfrac{13}{2} + 6 = 0$ ✓
③So $(2x - 1)$ is a factor. Compare coefficients: $\mathrm{f}(x) \equiv (2x - 1)\left(x^2 + Bx + C\right)$.
④Constant: $-C = 6 \Rightarrow C = -6$.  $x^2$: $2B - 1 = 1 \Rightarrow B = 1$.
⑤$\mathrm{f}(x) = (2x - 1)\left(x^2 + x - 6\right) = (2x - 1)(x + 3)(x - 2)$.
Worked Example 8 — Two unknown coefficients

$\mathrm{f}(x) = x^3 + ax^2 + bx - 6$. Both $(x - 1)$ and $(x + 2)$ are factors of $\mathrm{f}(x)$. Find $a$ and $b$, and the third factor.

①$\mathrm{f}(1) = 0$: $1 + a + b - 6 = 0$, so $a + b = 5$.
②$\mathrm{f}(-2) = 0$: $-8 + 4a - 2b - 6 = 0$, so $4a - 2b = 14$, i.e. $2a - b = 7$.
③Add the equations: $3a = 12$, so $a = 4$ and $b = 1$.
④$\mathrm{f}(x) = x^3 + 4x^2 + x - 6$. The constants of the factors multiply to $-6$: $(-1)(2)(k) = -6$ gives $k = 3$.
⑤Third factor $(x + 3)$. Check: $(x - 1)(x + 2)(x + 3) = x^3 + 4x^2 + x - 6$ ✓
Two unknowns need two equations — one from each factor. This is a classic place where simultaneous equations turn up inside another topic.
When the quadratic will not factorise. After dividing out one factor, check the discriminant $b^2 - 4ac$ of the quadratic. If it is negative, the cubic has only one real root; if it is positive but not a perfect square, use the quadratic formula for the other two roots (see Quadratic Functions).
6 Quick Reference

Degree

The highest power; a degree-$n$ polynomial has at most $n$ real roots.

Expanding

Every term times every term; degrees add, constants multiply.

Factorising order

Common factor → pattern → grouping → factor theorem.

Division

Long division (use $0x^2$ for gaps) or compare coefficients.

Factor theorem

$\mathrm{f}(a) = 0 \iff (x-a)$ is a factor.

$(ax - b)$

A factor exactly when $\mathrm{f}\!\left(\tfrac{b}{a}\right) = 0$.

Sign trap

For $(x + 3)$ substitute $x = -3$.

Unknown coefficients

One equation per known factor; solve simultaneously.

Leftover quadratic

Discriminant $< 0$ means the cubic has only one real root.

7 Practice Questions
Question 1

Expand and simplify $(x + 3)\left(2x^2 - x + 4\right)$.

▶ Show solution

$x\left(2x^2 - x + 4\right) = 2x^3 - x^2 + 4x$

$3\left(2x^2 - x + 4\right) = 6x^2 - 3x + 12$

Sum: $2x^3 + 5x^2 + x + 12$

Question 2

Expand $(x - 1)(x + 2)(x - 4)$, giving your answer in the form $x^3 + ax^2 + bx + c$.

▶ Show solution

$(x - 1)(x + 2) = x^2 + x - 2$

$\left(x^2 + x - 2\right)(x - 4) = x^3 - 4x^2 + x^2 - 4x - 2x + 8$

$= x^3 - 3x^2 - 6x + 8$, so $a = -3$, $b = -6$, $c = 8$.

Check: constant $(-1)(2)(-4) = 8$ ✓

Question 3

Factorise fully (a) $x^3 - 5x^2 + 6x$   (b) $4x^3 - 9x$.

▶ Show solution

(a) $x\left(x^2 - 5x + 6\right) = x(x - 2)(x - 3)$

(b) $x\left(4x^2 - 9\right) = x(2x - 3)(2x + 3)$ — a difference of two squares, since $4x^2 = (2x)^2$ and $9 = 3^2$.

Question 4

Divide $x^3 + 4x^2 - 7x - 10$ by $(x + 1)$, and hence factorise it fully.

▶ Show solution

$x^3 \div x = x^2$: $x^2(x + 1) = x^3 + x^2$; subtract to get $3x^2 - 7x$.

$3x^2 \div x = 3x$: $3x(x + 1) = 3x^2 + 3x$; subtract to get $-10x - 10$.

$-10x \div x = -10$: $-10(x + 1) = -10x - 10$; remainder $0$.

Quotient $x^2 + 3x - 10 = (x + 5)(x - 2)$.

$x^3 + 4x^2 - 7x - 10 = (x + 1)(x + 5)(x - 2)$

Question 5

$\mathrm{f}(x) = x^3 - 3x^2 - 4x + 12$. Use the factor theorem to show that $(x - 2)$ is a factor of $\mathrm{f}(x)$, and hence solve $\mathrm{f}(x) = 0$.

▶ Show solution

$\mathrm{f}(2) = 8 - 12 - 8 + 12 = 0$, so by the factor theorem $(x - 2)$ is a factor.

$\mathrm{f}(x) \equiv (x - 2)\left(x^2 + Bx + C\right)$: constant $-2C = 12 \Rightarrow C = -6$; $x^2$: $B - 2 = -3 \Rightarrow B = -1$.

$\mathrm{f}(x) = (x - 2)\left(x^2 - x - 6\right) = (x - 2)(x - 3)(x + 2)$

$x = -2,\; 2,\; 3$

Alternative: grouping gives $x^2(x - 3) - 4(x - 3) = (x - 3)(x^2 - 4)$ — the same answer.

Question 6

$(x + 3)$ is a factor of $\mathrm{f}(x) = 2x^3 + kx^2 - 7x + 6$. (a) Find $k$.   (b) Show that $\mathrm{f}(x) = 0$ has exactly one real root.

▶ Show solution

(a) $\mathrm{f}(-3) = 0$: $2(-27) + 9k + 21 + 6 = 0$

$-54 + 9k + 27 = 0 \Rightarrow 9k = 27 \Rightarrow k = 3$.

(b) $2x^3 + 3x^2 - 7x + 6 \equiv (x + 3)\left(2x^2 + Bx + C\right)$.

Constant: $3C = 6 \Rightarrow C = 2$.  $x^2$: $B + 6 = 3 \Rightarrow B = -3$.  Check $x$: $C + 3B = 2 - 9 = -7$ ✓

$\mathrm{f}(x) = (x + 3)\left(2x^2 - 3x + 2\right)$. The quadratic has discriminant $(-3)^2 - 4(2)(2) = 9 - 16 = -7 < 0$, so it has no real roots.

The only real root is $x = -3$.

Question 7

Solve $2x^3 - 5x^2 - 4x + 3 = 0$.

▶ Show solution

Try $x = 1$: $2 - 5 - 4 + 3 = -4$.  Try $x = -1$: $-2 - 5 + 4 + 3 = 0$ ✓, so $(x + 1)$ is a factor.

$2x^3 - 5x^2 - 4x + 3 = (x + 1)\left(2x^2 - 7x + 3\right)$ (constant: $C = 3$; $x^2$: $B + 2 = -5 \Rightarrow B = -7$).

$2x^2 - 7x + 3 = (2x - 1)(x - 3)$

$x = -1,\; \tfrac12,\; 3$

Question 8

$\mathrm{f}(x) = 2x^3 + px^2 + qx - 4$. Given that $(x - 2)$ is a factor of $\mathrm{f}(x)$ and that $\mathrm{f}(-1) = -6$, find $p$ and $q$. Hence factorise $\mathrm{f}(x)$ as far as possible.

▶ Show solution

$\mathrm{f}(2) = 0$: $16 + 4p + 2q - 4 = 0 \Rightarrow 2p + q = -6$.

$\mathrm{f}(-1) = -6$: $-2 + p - q - 4 = -6 \Rightarrow p - q = 0$.

So $p = q$, and $3p = -6$ gives $p = -2$, $q = -2$.

$\mathrm{f}(x) = 2x^3 - 2x^2 - 2x - 4 = 2\left(x^3 - x^2 - x - 2\right) = 2(x - 2)\left(x^2 + x + 1\right)$

$x^2 + x + 1$ has discriminant $1 - 4 = -3 < 0$, so it does not factorise further.

Question 9

Show that $(3x - 2)$ is a factor of $\mathrm{f}(x) = 3x^3 - 2x^2 - 12x + 8$, and hence factorise $\mathrm{f}(x)$ completely.

▶ Show solution

$3x - 2 = 0$ when $x = \tfrac23$.

$\mathrm{f}\!\left(\tfrac23\right) = 3\left(\tfrac{8}{27}\right) - 2\left(\tfrac49\right) - 8 + 8 = \tfrac89 - \tfrac89 = 0$ ✓, so $(3x - 2)$ is a factor.

Grouping: $x^2(3x - 2) - 4(3x - 2) = (3x - 2)\left(x^2 - 4\right)$

$\mathrm{f}(x) = (3x - 2)(x - 2)(x + 2)$

Question 10

A cuboid has edges of length $x$ cm, $(x + 2)$ cm and $(x + 5)$ cm, and its volume is $120$ cm$^3$.

(a) Show that $x^3 + 7x^2 + 10x - 120 = 0$.   (b) Show that $(x - 3)$ is a factor of $x^3 + 7x^2 + 10x - 120$.   (c) Show that $x = 3$ is the only real solution, and state the dimensions of the cuboid.   (d) Find the surface area of the cuboid.

▶ Show solution

(a) $x(x + 2)(x + 5) = 120$. Expanding: $x\left(x^2 + 7x + 10\right) = x^3 + 7x^2 + 10x$, so $x^3 + 7x^2 + 10x - 120 = 0$.

(b) With $\mathrm{g}(x) = x^3 + 7x^2 + 10x - 120$: $\mathrm{g}(3) = 27 + 63 + 30 - 120 = 0$, so $(x - 3)$ is a factor.

(c) $\mathrm{g}(x) \equiv (x - 3)\left(x^2 + Bx + C\right)$: constant $-3C = -120 \Rightarrow C = 40$; $x^2$: $B - 3 = 7 \Rightarrow B = 10$. Check $x$: $C - 3B = 40 - 30 = 10$ ✓

$x^2 + 10x + 40$ has discriminant $100 - 160 = -60 < 0$, so no further real roots. Hence $x = 3$ only.

The cuboid is $3$ cm by $5$ cm by $8$ cm. (Check: $3 \times 5 \times 8 = 120$ ✓)

(d) Surface area $= 2(3 \times 5 + 3 \times 8 + 5 \times 8) = 2(15 + 24 + 40) = 158$ cm$^2$.

Polynomials and the Factor Theorem · A Level Maths Year 1 · Created with MathJax