Cubic Roots after Division

From the factorised form to the roots of f(x) = 0, exact answers

In the exam for Edexcel International AS Maths · 5 real past-paper questions below

Roots from the factors

You can already divide a cubic by a factor and write it as a bracket times a quadratic. AS questions then ask the next thing: solve $f (x) = 0$ .

Rule

A product is zero only when one of its factors is zero. So set each bracket equal to 0 and solve it.

Worked example

1
Given that $x + 2$ is a factor of $f (x) = 2 x^{3} + 2 x^{2} - 5 x - 2$ , solve $f (x) = 0$ . Give exact answers.
2
$f (x) = (x + 2) (2 x^{2} - 2 x - 1)$ — Divide by x + 2 (or compare coefficients). The remainder is 0.
3
$x + 2 = 0 \Rightarrow x = - 2$ — First bracket equal to zero.
4
$2 x^{2} - 2 x - 1 = 0$ — Second bracket. It does not factorise, so use the formula.
5
$x = \frac{2 \pm 4 + 8}{4} = \frac{2 \pm 2 3}{4}$ — a = 2, b = −2, c = −1, so b² − 4ac = 12.
6
$x = \frac{1 \pm 3}{2}$ — √12 = 2√3, then cancel 2. Answers: −2 and (1 ± √3)/2.

Common mistake

Giving $x = 1.37$ and $x = - 0.37$ when the question says “exact”. Decimals lose the accuracy mark. Leave surds as surds.

Common mistake

Answering with the brackets, such as “ $x + 2$ ”. A root is a number: $x = - 2$ .

Key idea

Factorise, set each bracket to zero, and give each root as a number, exact unless told otherwise.

How many real roots?

Often the question is not “solve” but “how many real roots?” or “show that there is only one real root”. The quadratic factor decides it, through its discriminant $b^{2} - 4 a c$ .

Formula

b^{2} - 4 a c

Greater than 0: the quadratic gives two different real roots.
Equal to 0: one repeated root. It counts once.
Less than 0: no real roots. The square root in the formula has nothing real to give.

Memory trick

Think of the discriminant as the number under the square root in the formula. Negative under the root, no real answers: the formula “breaks”.

Worked example

1
$f (x) = 3 x^{3} + x^{2} + 4 x - 4$ . Given that $3 x - 2$ is a factor, show that $f (x) = 0$ has only one real root.
2
$f (x) = (3 x - 2) (x^{2} + x + 2)$ — Divide by 3x − 2. Remainder 0.
3
$b^{2} - 4 a c = 1^{2} - 4 (1) (2) = - 7 < 0$ — Discriminant of the quadratic factor. Say that it is negative.
4
$x^{2} + x + 2 = 0 has no real roots$ — So the quadratic adds nothing.
5
$only real root: x = \frac{2}{3}$ — It comes from 3x − 2 = 0. Write the conclusion.

Marks come for the quadratic factor, then for testing it with $b^{2} - 4 a c$ (or the formula, or completing the square). The last mark needs the right value, “< 0”, “no real roots” and the conclusion. “No roots” with no reason scores nothing.

Common mistake

Trying to factorise $x^{2} + x + 2$ and giving up. A failed factorisation proves nothing. Only the discriminant does.

Common mistake

Counting brackets, not roots. A repeated root counts once. In $(x - 2) (x^{2} - 5 x + 6)$ the discriminant is positive, but $x^{2} - 5 x + 6 = (x - 2) (x - 3)$ . The roots are 2 and 3: two distinct solutions, not three.

Key idea

Discriminant less than 0: only the linear factor gives a root. Then count the different numbers, not the brackets.

Expand first, then divide

Some questions give the cubic in a disguised form: a bracket times a quadratic, plus a number. Two things follow from that shape.

Formula

f (x) = (x - a) q (x) + R

When you divide $f (x)$ by $x - a$ , the remainder is $R$ . You can just read it off.
But $x - a$ is not a factor (unless R = 0), so $q (x)$ is not the quadratic you need. To use a different factor, expand first.

Worked example

1
$f (x) = (x + 1) (x^{2} + k x - 4) - 6$ , where k is a constant. Find the remainder when $f (x)$ is divided by $x + 1$ . Then, given that $x - 2$ is a factor, find k and the number of real solutions of $f (x) = 0$ .
2
$R = - 6$ — Put x = −1: the first part is 0, so −6 is left.
3
$f (2) = 3 (4 + 2 k - 4) - 6 = 0$ — Factor theorem: x − 2 is a factor, so f(2) = 0.
4
$6 k - 6 = 0 \Rightarrow k = 1$ — Simplify and solve.
5
$f (x) = x^{3} + 2 x^{2} - 3 x - 4 - 6 = x^{3} + 2 x^{2} - 3 x - 10$ — Expand (x + 1)(x² + x − 4), then subtract 6.
6
$f (x) = (x - 2) (x^{2} + 4 x + 5)$ — Now divide the full cubic by x − 2.
7
$b^{2} - 4 a c = 16 - 20 = - 4 < 0$ — The new quadratic has no real roots.
8
$one real solution, x = 2$ — Give the number and the reason.

Common mistake

Testing $x^{2} + x - 4$ , the quadratic in the given form. It goes with $x + 1$ , which is not a factor. Its discriminant is 17, which wrongly suggests three solutions. It earns no marks.

Key idea

Read the remainder straight off the given form. To factorise, expand to a plain cubic first, then divide by the real factor.

Cubics from logs: check roots

A log equation can turn into a cubic. You solve the cubic as usual, but the log equation is the real question. Logs only accept positive inputs, so some roots must go.

Worked example

1
Solve $lo g_{3} (x + 1) + 2 lo g_{3} x = 1 + lo g_{3} (8 x - 12)$ , using algebra and showing each step.
2
$1 = lo g_{3} 3$ — Write the number as a log in the same base.
3
$lo g_{3} (x^{2} (x + 1)) = lo g_{3} 3 (8 x - 12)$ — Power law, then sum law, on each side.
4
$x^{2} (x + 1) = 24 x - 36$ — Same log on both sides, so the insides are equal.
5
$x^{3} + x^{2} - 24 x + 36 = 0$ — Expand and collect.
6
$f (2) = 8 + 4 - 48 + 36 = 0$ — Factor theorem: x − 2 is a factor.
7
$(x - 2) (x - 3) (x + 6) = 0$ — Divide to get x² + 3x − 18, then factorise.
8
$x = - 6 : x + 1 = - 5$ — log₃ of a negative number does not exist. Reject x = −6.
9
$x = 2 : 3, 2, 4 x = 3 : 4, 3, 12$ — All log inputs positive each time. Answers: x = 2, x = 3.

Rule

Check each root in the original equation. Every log input, here $x + 1$ , $x$ and $8 x - 12$ , must be greater than 0.

Marks: the last accuracy mark is lost if a rejected value is still in your answer. “Showing each step of your working” means roots read off a calculator score nothing; the division and the quadratic must be on the page.

Common mistake

Checking only in the cubic. $x = - 6$ is a true root of the cubic. It fails only in the log equation, where $lo g_{3} (- 6)$ does not exist.

Key idea

The cubic finds candidates. The log equation decides which ones survive.

Which method? Read the question

The wording tells you what to do with the factorised cubic.

You see	You do
“Solve $f (x) = 0$ ”, exact	Each bracket = 0; formula for the quadratic; leave surds
“Show that … only one real root”	Quadratic factor: $b^{2} - 4 a c$ , state < 0, conclude
“Find the number of real solutions”	Discriminant, then count different roots only
$f (x) = (x - a) q (x) + R$	Remainder for $x - a$ is R; expand before using another factor
Cubic from a log equation	Solve, then check every log input is positive

Common mistake

Stopping at the factorisation. If the question asks for roots, a number, or a reason, give it in a final sentence.

Key idea

Factorise, test the quadratic, count the different roots, and check each one against the question you started with.

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Real past-paper questions on cubic roots after division

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