Polynomials and the Factor Theorem – OCR FSMQ Additional Maths

1 What a Polynomial Is

A polynomial is a sum of terms of the form $kx^n$ where every index $n$ is a non-negative whole number. The degree is the highest power present.

Polynomial	Degree	Name
$3x + 1$	$1$	Linear
$2x^2 - x + 5$	$2$	Quadratic
$x^3 - 4x$	$3$	Cubic
$x^4 + 2x^2 - 7$	$4$	Quartic

These are not polynomials: $\dfrac{1}{x} = x^{-1}$ (negative index), $\sqrt{x} = x^{1/2}$ (fractional index), $2^x$ (variable in the index).

A degree-$n$ polynomial has at most $n$ roots. So a cubic equation has one, two or three solutions — never more. Use this as a check on your working.

2 Adding, Subtracting and Multiplying

AL3 asks for all four operations with polynomials. Addition and subtraction just collect like terms; multiplication means every term times every term.

Worked Example 1 — Addition and subtraction

$\mathrm{p}(x) = 2x^3 - x + 4$ and $\mathrm{q}(x) = x^3 + 3x^2 - 5$. Find $\mathrm{p}(x) + \mathrm{q}(x)$ and $\mathrm{p}(x) - \mathrm{q}(x)$.

①Sum: $(2+1)x^3 + 3x^2 - x + (4-5) = 3x^3 + 3x^2 - x - 1$

②Difference: $(2-1)x^3 - 3x^2 - x + (4+5)$

③$= x^3 - 3x^2 - x + 9$

Subtracting changes the sign of every term of $\mathrm{q}(x)$, including the $-5$, which becomes $+5$.

Worked Example 2 — Multiplying two brackets

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

①$x \times \left(x^2 + 3x - 1\right) = x^3 + 3x^2 - x$

②$-2 \times \left(x^2 + 3x - 1\right) = -2x^2 - 6x + 2$

③Add: $x^3 + (3-2)x^2 + (-1-6)x + 2$

④$= x^3 + x^2 - 7x + 2$

Degree check: degree $1 \times$ degree $2$ must give degree $3$, and the constant term must be $(-2)(-1) = 2$ ✓

3 Dividing Polynomials

Once you know one factor, division finds the rest. There are two reliable methods.

Method A — long division

Divide the leading terms, multiply back, subtract, bring down the next term, repeat

Worked Example 3 — Long division

Divide $x^3 - 4x^2 + x + 6$ by $(x + 1)$.

①$x^3 \div x = x^2$. Multiply: $x^2(x+1) = x^3 + x^2$. Subtract: $-5x^2 + x + 6$.

②$-5x^2 \div x = -5x$. Multiply: $-5x(x+1) = -5x^2 - 5x$. Subtract: $6x + 6$.

③$6x \div x = 6$. Multiply: $6(x+1) = 6x + 6$. Subtract: $0$.

④Quotient $x^2 - 5x + 6$, remainder $0$.

⑤So $x^3 - 4x^2 + x + 6 = (x+1)\left(x^2 - 5x + 6\right)$.

Method B — comparing coefficients

Write $x^3 + bx^2 + cx + d \equiv (x - a)\left(x^2 + px + q\right)$ and match each power

Worked Example 4 — Comparing coefficients

Given that $(x - 2)$ is a factor, write $x^3 + x^2 - 8x + 4$ as $(x-2)\left(x^2 + px + q\right)$.

①Expand the right side: $x^3 + px^2 + qx - 2x^2 - 2px - 2q$

②$= x^3 + (p-2)x^2 + (q - 2p)x - 2q$

③Constant: $-2q = 4$, so $q = -2$.

④$x^2$: $p - 2 = 1$, so $p = 3$.

⑤Check the $x$ term: $q - 2p = -2 - 6 = -8$ ✓

⑥$x^3 + x^2 - 8x + 4 = (x-2)\left(x^2 + 3x - 2\right)$

The fastest route: for a cubic with leading coefficient $1$ and factor $(x-a)$, the quadratic is $x^2 + px + q$ where $-aq$ is the constant term. Start with the constant and the $x^2$ coefficient, then use the $x$ coefficient as a check.

4 The Factor Theorem

The factor theorem (AL4)

If $\mathrm{f}(a) = 0$ then $(x - a)$ is a factor of $\mathrm{f}(x)$
and conversely, if $(x-a)$ is a factor then $\mathrm{f}(a) = 0$

Why it works. If $(x-a)$ is a factor then $\mathrm{f}(x) = (x-a) \times \mathrm{g}(x)$. Putting $x = a$ makes the first bracket zero, so the whole thing is zero.

Mind the sign. To test the factor $(x + 3)$ you substitute $x = -3$, because $x + 3 = 0$ when $x = -3$. The factor $(x - 3)$ needs $x = +3$.

Which values to try

Test the factors of the constant term, positive and negative

Worked Example 5 — Fully factorising a cubic

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

①The constant is $6$, so try $x = \pm1, \pm2, \pm3, \pm6$.

②$\mathrm{f}(1) = 1 - 2 - 5 + 6 = 0$ ✓ so $(x-1)$ is a factor.

③Divide: $x^3 - 2x^2 - 5x + 6 = (x-1)\left(x^2 - x - 6\right)$

④$x^2 - x - 6 = (x-3)(x+2)$

⑤$\mathrm{f}(x) = (x-1)(x-3)(x+2)$

⑥$\mathrm{f}(x) = 0$ when $x = 1$, $x = 3$ or $x = -2$.

Check without expanding: the three constants multiply to $(-1)(-3)(2) = 6$ ✓, matching the constant term of $\mathrm{f}(x)$.

Worked Example 6 — Finding an unknown coefficient

$(x - 3)$ is a factor of $\mathrm{f}(x) = 2x^3 - 5x^2 + kx - 6$. Find $k$.

①The factor theorem gives $\mathrm{f}(3) = 0$.

②$2(27) - 5(9) + 3k - 6 = 0$

③$54 - 45 + 3k - 6 = 0$

④$3 + 3k = 0$, so $k = -1$.

Worked Example 7 — A repeated factor

Factorise $\mathrm{f}(x) = x^3 - 3x^2 + 4$ and describe the shape of its graph near each root.

①$\mathrm{f}(-1) = -1 - 3 + 4 = 0$, so $(x+1)$ is a factor.

②Dividing gives $(x+1)\left(x^2 - 4x + 4\right)$.

③$x^2 - 4x + 4 = (x-2)^2$

④$\mathrm{f}(x) = (x+1)(x-2)^2$

⑤At $x = -1$ the graph crosses the axis. At $x = 2$ the factor is squared, so the graph touches the axis and turns back — $x=2$ is a stationary point.

A squared factor always means the curve touches rather than crosses. That is a free piece of information for sketching questions (CG4).

5 When There Is a Remainder

If $\mathrm{f}(a) \neq 0$ then $(x-a)$ is not a factor — and the value $\mathrm{f}(a)$ is exactly the remainder when you divide $\mathrm{f}(x)$ by $(x-a)$.

Worked Example 8 — A non-zero remainder

Find the remainder when $x^3 + 2x^2 - x + 1$ is divided by $(x - 2)$.

①Substitute $x = 2$: $8 + 8 - 2 + 1 = 15$

②The remainder is $15$, so $(x-2)$ is not a factor.

③Equivalently $x^3 + 2x^2 - x + 1 = (x-2)\left(x^2 + 4x + 7\right) + 15$.

A single substitution is far quicker than long division when all you need is the remainder.

6 Quick Reference

Degree

Highest power; at most that many roots.

Subtracting

Change the sign of every term.

Multiplying

Degrees add; constants multiply.

Factor theorem

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

Sign trap

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

What to try

Factors of the constant term, $\pm$.

Remainder

Dividing by $(x-a)$ leaves $\mathrm{f}(a)$.

Squared factor

Graph touches the axis there.

Division

Long division, or compare coefficients.

Final check

Roots' constants must multiply to the constant term.

7 Practice Questions

Question 1

Expand and simplify $(x + 4)\left(x^2 - 2x + 5\right)$.

▶ Show solution

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

$4\left(x^2 - 2x + 5\right) = 4x^2 - 8x + 20$

Sum: $x^3 + 2x^2 - 3x + 20$

Question 2

Show that $(x - 1)$ is a factor of $x^3 + 2x^2 - 5x + 2$.

▶ Show solution

$\mathrm{f}(1) = 1 + 2 - 5 + 2 = 0$

Since $\mathrm{f}(1) = 0$, by the factor theorem $(x-1)$ is a factor.

Question 3

Find the remainder when $2x^3 - x^2 + 4$ is divided by $(x + 1)$.

▶ Show solution

Substitute $x = -1$:

$2(-1) - (1) + 4 = -2 - 1 + 4 = 1$

The remainder is $1$.

Question 4

Divide $x^3 - 7x + 6$ by $(x - 1)$.

▶ Show solution

Note there is no $x^2$ term, so treat it as $x^3 + 0x^2 - 7x + 6$.

$x^3 \div x = x^2$; subtract $x^3 - x^2$ to leave $x^2 - 7x + 6$.

$x^2 \div x = x$; subtract $x^2 - x$ to leave $-6x + 6$.

$-6x \div x = -6$; subtract $-6x + 6$ to leave $0$.

Quotient $x^2 + x - 6$, remainder $0$.

Question 5

Factorise $x^3 + 6x^2 + 11x + 6$ completely.

▶ Show solution

Try $x = -1$: $-1 + 6 - 11 + 6 = 0$ ✓ so $(x+1)$ is a factor.

Dividing: $(x+1)\left(x^2 + 5x + 6\right)$

$x^2 + 5x + 6 = (x+2)(x+3)$

$= (x+1)(x+2)(x+3)$

Question 6

$(x + 2)$ is a factor of $x^3 + ax^2 + 3x - 2$. Find $a$.

▶ Show solution

$\mathrm{f}(-2) = 0$:

$-8 + 4a - 6 - 2 = 0$

$4a - 16 = 0$, so $a = 4$.

Question 7

Solve $x^3 - 4x^2 - 3x + 18 = 0$.

▶ Show solution

Try factors of $18$. $\mathrm{f}(3) = 27 - 36 - 9 + 18 = 0$ ✓

Dividing by $(x-3)$: $x^2 - x - 6 = (x-3)(x+2)$

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

$x = 3$ (a repeated root) or $x = -2$.

Question 8

$\mathrm{f}(x) = x^3 + px^2 + qx + 12$ has factors $(x-1)$ and $(x-2)$. Find $p$ and $q$.

▶ Show solution

$\mathrm{f}(1) = 0$: $1 + p + q + 12 = 0$, so $p + q = -13$.

$\mathrm{f}(2) = 0$: $8 + 4p + 2q + 12 = 0$, so $4p + 2q = -20$, i.e. $2p + q = -10$.

Subtracting: $p = -10 - (-13) = 3$.

Then $q = -13 - 3 = -16$.

Check: the third root must satisfy $1 \times 2 \times r = -12$, so $r = -6$, and $1 + 2 - 6 = -3 = -p$ ✓

Question 9

Explain why $x^3 + x + 1$ has no factor of the form $(x - a)$ with $a$ a whole number.

▶ Show solution

Any such factor requires $a$ to divide the constant term $1$, so the only candidates are $a = 1$ and $a = -1$.

$\mathrm{f}(1) = 1 + 1 + 1 = 3 \neq 0$

$\mathrm{f}(-1) = -1 - 1 + 1 = -1 \neq 0$

Neither gives zero, so there is no linear factor with integer $a$.

(The cubic does still have one real root — it just is not a whole number. Finding it is a job for numerical methods.)

Question 10

An open box is made from a square sheet of card of side $12$ cm by cutting a square of side $x$ cm from each corner and folding up the sides.

(a) Show that the volume is $V = x(12-2x)^2$. (b) Expand this to a cubic in $x$. (c) Show that $V = 128$ has a solution $x = 2$, and find the other positive solution, giving it to 2 decimal places. (d) State the range of values of $x$ that make physical sense.

▶ Show solution

(a) Cutting $x$ from both ends of each side leaves a base of $(12 - 2x)$ by $(12-2x)$, and the height of the folded box is $x$.

$V = x(12-2x)^2$

(b) $(12-2x)^2 = 144 - 48x + 4x^2$

$V = 144x - 48x^2 + 4x^3$

(c) At $x = 2$: $V = 2(12-4)^2 = 2 \times 64 = 128$ ✓

So solve $4x^3 - 48x^2 + 144x - 128 = 0$, i.e. $x^3 - 12x^2 + 36x - 32 = 0$.

Since $x=2$ is a root, divide by $(x-2)$: $x^2 - 10x + 16$.

$x = \dfrac{10 \pm \sqrt{100 - 64}}{2} = \dfrac{10 \pm 6}{2} = 8 \text{ or } 2$.

So the roots are $x = 2$ (twice) and $x = 8$. The other positive solution is $x = 8.00$.

(d) Need $x > 0$ for a box to exist, and $12 - 2x > 0$ for a base to exist, so $x < 6$.

Valid range: $0 < x < 6$. This rules out $x = 8$ — it satisfies the algebra but not the geometry, which is exactly the kind of interpretation AO3 rewards.

Polynomials and the Factor Theorem (AL3–AL4) · OCR FSMQ Additional Maths · Created with MathJax