Cubic Equations and Curve Meetings

No number term: take out x first, then solve the quadratic (no factor theorem needed on P1)

In the exam for Edexcel International AS Maths · 6 real past-paper questions below

No number term: take out x

You can already solve a cubic by finding one root and dividing. AS Pure 1 questions are kinder: the cubic usually has no number term. Every term has an x in it, so x itself is a factor.

Rule

No number term? Take out x. What is left is a quadratic, and you already know how to solve that.

Worked example

1
Solve $x^{2} (3 x - 1) = 10 x$ , using algebra.
2
$3 x^{3} - x^{2} - 10 x = 0$ — Multiply out the bracket and bring everything to one side.
3
$x (3 x^{2} - x - 10) = 0$ — Every term has an x, so take it out.
4
$x (3 x + 5) (x - 2) = 0$ — Factorise the quadratic: (3x + 5)(x − 2) = 3x² − x − 10.
5
$x = 0, x = - \frac{5}{3}, x = 2$ — Each factor equal to zero. The lone x gives x = 0.

Common mistake

Dividing both sides by x. That throws away the root $x = 0$ , and the final accuracy mark needs all three roots.

Memory trick

Take x out, never divide it out. Dividing by x is like tearing a page out of the answer.

A “factorise fully” question works the same way, and the quadratic left may be a perfect square: $x^{3} - 8 x^{2} + 16 x = x (x - 4)^{2}$ .

Rule

“Using algebra” or “show detailed reasoning” means show the factorised line before the roots. Roots straight from a calculator score 0.

Key idea

No number term means x is a factor. One step turns the cubic into a quadratic.

Hence: swap the letter back

Next, a “hence” part gives a longer equation built from the same cubic. Spot what plays the part of x, use your roots, then solve back for the new letter.

Say you have just found that $3 x^{3} - x^{2} - 10 x = 0$ has roots $0, - \frac{5}{3}, 2$ .

Worked example

1
Hence solve $3 (y - 1)^{\frac{3}{2}} - (y - 1) - 10 (y - 1)^{\frac{1}{2}} = 0$ .
2
$x = (y - 1)^{\frac{1}{2}}$ — Spot the swap: then x³ is (y − 1)^(3/2) and x² is y − 1.
3
$(y - 1)^{\frac{1}{2}} = 0, - \frac{5}{3}, 2$ — Set the swap equal to each root from before.
4
$(y - 1)^{\frac{1}{2}} = - \frac{5}{3}$ — Reject: a square root is never negative.
5
$y - 1 = 0 \Rightarrow y = 1$ — Square both sides, then add 1.
6
$y - 1 = 4 \Rightarrow y = 5$ — Square 2 to get 4, then add 1. Answers: y = 1, y = 5.

Example

Cube roots are different. For $3 y - y^{\frac{2}{3}} - 10 y^{\frac{1}{3}} = 0$ , the swap is $x = y^{\frac{1}{3}}$ . Cube each root: $y = 0, - \frac{125}{27}, 8$ . Keep the negative one: $(- \frac{5}{3})^{3}$ is a real number.

Worked example

1
Hence solve $3 (4^{3 z}) - 4^{2 z} - 10 (4^{z}) = 0$ .
2
$x = 4^{z}$ — Here 4^(3z) = (4^z)³ and 4^(2z) = (4^z)².
3
$4^{z} = 0 or - \frac{5}{3}$ — Reject both: a power of 4 is always positive.
4
$4^{z} = 2 = 4^{\frac{1}{2}} \Rightarrow z = \frac{1}{2}$ — Write 2 as a power of 4. The only answer.

Marks: one for setting the swap equal to a root, one for a full method back to the new letter, and accuracy marks for the answers. A rejected value left in your final list costs the last mark.

Common mistake

Squaring $(y - 1)^{\frac{1}{2}} = - \frac{5}{3}$ to get $y = \frac{34}{9}$ . Check it: the square root of $\frac{25}{9}$ is $+ \frac{5}{3}$ , not $- \frac{5}{3}$ . Reject it and say why.

Key idea

Find the swap, use the roots you have, then ask of each one: can the swap really equal this?

Build a cubic from its roots

Sometimes you work backwards: you know where a cubic crosses the x-axis and one other point, and must find its equation.

Formula

y = A (x - p) (x - q) (x - r)

$p, q, r$ — the roots. Each bracket is x minus a root, so a root of −3 gives (x + 3).
$A$ — a number that stretches the curve. Find it from the extra point.

Why the A? $(x + 3) (x - 1) (x - 4)$ and $5 (x + 3) (x - 1) (x - 4)$ have the same roots. The extra point picks the right one.

Worked example

1
A cubic crosses the x-axis at $- 3$ , $1$ and $4$ , and passes through $(2, 15)$ . Find it in the form $y = a x^{3} + b x^{2} + c x + d$ .
2
$y = A (x + 3) (x - 1) (x - 4)$ — One bracket for each root. Watch the signs.
3
$15 = A (5) (1) (- 2) = - 10 A$ — Put in x = 2, y = 15.
4
$A = - \frac{3}{2}$ — Divide by −10.
5
$(x + 3) (x - 1) (x - 4) = x^{3} - 2 x^{2} - 11 x + 12$ — Expand two brackets, then times the third.
6
$y = - \frac{3}{2} x^{3} + 3 x^{2} + \frac{33}{2} x - 18$ — Multiply every term by A.

Example

If the curve touches the axis at a root, that bracket is squared. Touching at the origin and crossing at 5, through $(6, 18)$ : $y = A x^{2} (x - 5)$ , so $18 = 36 A$ , $A = \frac{1}{2}$ , $y = \frac{1}{2} x^{3} - \frac{5}{2} x^{2}$ .

When the answer must be in the form ax³ + bx² + cx + d, marks go for the brackets, for using the point to find A, for A, for expanding, and for the final values. Check: put x = 2 into your answer. You should get 15.

Common mistake

Leaving out A and writing $y = (x + 3) (x - 1) (x - 4)$ . That curve passes through $(2, - 10)$ , not $(2, 15)$ .

Key idea

Roots give the brackets. The extra point gives A. Then expand.

Where two curves meet

Two curves meet where they have the same y at the same x. So set the two expressions equal. With a cubic and a quadratic, you get a cubic, often with no number term.

Rule

Equate, bring everything to one side, take out x, solve the quadratic. Then find y for every x.

Worked example

1
Find where $y = x^{3} - 2 x$ meets $y = 4 x - x^{2}$ . Use algebra and show each stage.
2
$x^{3} - 2 x = 4 x - x^{2}$ — Same y at a meeting point.
3
$x^{3} + x^{2} - 6 x = 0$ — Collect everything on one side.
4
$x (x + 3) (x - 2) = 0$ — Take out x, then factorise x² + x − 6.
5
$x = 0, - 3, 2$ — Each factor equal to zero.
6
$y = 0, - 21, 4$ — Put each x into the simpler curve, y = 4x − x².
7
$(0, 0), (- 3, - 21), (2, 4)$ — Give the points as coordinates.

If the question says the curves meet at the origin and asks only for another point, still show the factor x, then pick the x the question wants, such as the negative one.

Common mistake

Cancelling x from both sides at the start. You lose the meeting point at the origin, and the mark scheme looks for the factor x.

Common mistake

Stopping at the x-values. A point needs a y too. Check it in both equations: at $x = - 3$ each gives $- 21$ .

Key idea

Ray tracing in films and games finds where a light ray meets a curved surface in just this way: set two equations equal and solve. A computer does it millions of times per frame.

Key idea

Meeting means equal. Equate, factor, then find both coordinates.

Which method? Read the question

The wording tells you which move to make.

You see	You do
A cubic with no number term	Take out x; solve the quadratic; keep x = 0
“Factorise fully”	Take out x; the quadratic may be a perfect square
“Hence” with $(y + k)^{\frac{1}{2}}$	Swap = each root; reject negatives; square back
“Hence” with $y^{\frac{1}{3}}$	Cube each root; negatives are fine
“Hence” with $a^{z}$	Swap = each root; reject 0 and negatives; write as a power of a
Roots and one point given	$y = A (x - p) (x - q) (x - r)$ ; use the point for A; expand
Two curves meet	Equate; take out x; solve; find each y

Common mistake

Ending with a list that still holds a value you should have rejected. Cross it out and give a reason, such as “a square root cannot be negative”.

Key idea

Take out x, swap back with care, and always finish what the question asks for: roots, values, or points.

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