Solving cubic equations · Polynomials & algebraic division

Roots and factors

If $p$ , $q$ and $r$ are the roots of a cubic equation, then $(x - p)$ , $(x - q)$ and $(x - r)$ are its factors. On a graph, the roots are where the curve crosses the $x$ -axis.

Three rootsy = (x − p)(x − q)(x − r) crosses the x-axis at p, q and r

A repeated rootA squared factor (x − a)² makes the curve touch the axis at a

Try it

A cubic from its rootsMove the roots

−1, 2, 4roots8constant term

p = −1q = 2r = 4

y = (x + 1)(x − 2)(x − 4) = x³ − 5x² + 2x + 8The curve crosses the x-axis at each root, so x³ − 5x² + 2x + 8 = 0 has roots −1, 2, 4. The constant term is −pqr = 8, and every whole-number root divides it: that is where to look for a first root.

Move the roots and watch the brackets multiply out. Make two roots equal to see the curve touch the axis. Notice that the constant term is always $-pqr$ .

Solving a cubic

Find one root by trial. A whole-number root must divide the constant term, so try $\pm1$ , $\pm2$ , … until $f(a) = 0$ .
Divide by $(x - a)$ (long division, or by matching coefficients) to get a quadratic.
Solve the quadratic by factorising or the formula.

Some cubics factorise by grouping: take a common factor from the first two terms and from the last two.

Worked example · JAMB 2004

JAMB 2004 · UME · Q11

Find the values of $x$ where the curve $y = x^3 + 2x^2 - 5x - 6$ crosses the $x$ -axis.

Crossing the x-axis
The curve crosses the $x$ -axis where $y = 0$ , so solve $x^3 + 2x^2 - 5x - 6 = 0$ .
Think first. What is y there?
A first root
$(-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0$ , so $x = -1$ is a root and $x + 1$ is a factor.
Think first. Try x = −1: the constant is −6, so ±1, ±2, ±3, ±6 are the candidates.
Divide
$x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6)$ .
Think first. Divide by x + 1.
Factorise the quadratic
$x^2 + x - 6 = (x + 3)(x - 2)$ . So the curve crosses at $x = 2$ , $-1$ and $-3$ : option C.
Think first. Two numbers multiplying to −6 and adding to 1.

From roots to the polynomial

To build a polynomial with given roots, turn each root into a factor and multiply out. The root $-2$ gives the factor $(x + 2)$ .

Worked example · JAMB 2017

JAMB 2017 · UTME · Q16

A polynomial in $x$ whose zeros are $-2$ , $-1$ and $3$ is

Roots to factors
$-2$ , $-1$ and $3$ give $(x + 2)$ , $(x + 1)$ and $(x - 3)$ .
Think first. Which factor does each zero give?
Multiply two
$(x + 2)(x + 1) = x^2 + 3x + 2$ .
Think first. (x + 2)(x + 1) = ?
Multiply by the third
$x^3 - 3x^2 + 3x^2 - 9x + 2x - 6$ , which is $x^3 - 7x - 6$ : option D.
Think first. (x² + 3x + 2)(x − 3) = ?

Sum and difference of two cubes

Two cubes factorise with a linear bracket and a quadratic one:

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

same sign as the question, then the opposite sign, then always +

Sum and difference of cubesA linear bracket times a quadratic bracket, with the signs as marked

Your turn

JAMB 2002 · UME · Q39

Polynomials & algebraic division

Solve for $x$ in the equation $x^3 - 5x^2 - x + 5 = 0$ .

Worked solution (try it first)

Group the terms in pairs: $x^2(x - 5) - 1(x - 5) = 0$ .
Take out the common bracket: $(x - 5)(x^2 - 1) = 0$ , and $x^2 - 1 = (x - 1)(x + 1)$ .
So $x = 5$ , $x = 1$ or $x = -1$ , option D.

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