The remainder and factor theorems · Polynomials & algebraic division

The remainder theorem

When $f(x)$ is divided by $x - a$ , the remainder is a number $R$ , and $f(x) = (x - a)\,q(x) + R$ . Put $x = a$ : the bracket becomes 0, so $f(a) = R$ .

The remainder when $f(x)$ is divided by $x - a$ is $f(a)$ .

So you can find a remainder without dividing: just substitute. On a graph, $f(a)$ is the height of the curve at $x = a$ .

Remainder = f(a)The height of y = f(x) at x = a is the remainder on dividing by x − a

Watch the sign. Dividing by $x + 3$ means $x - (-3)$ , so substitute $x = -3$ . Dividing by $2x - 1$ means substituting the value that makes it zero, $x = \frac12$ .

The factor theorem

A factor divides exactly, with remainder 0. So $x - a$ is a factor of $f(x)$ exactly when $f(a) = 0$ .

On the graph, $f(a) = 0$ means the curve crosses the $x$ -axis at $x = a$ .

Factor: f(a) = 0The curve meets the x-axis at x = a, so x − a is a factor

Try it

The remainder theoremSlide a along the x-axis

(x − 2)divide by−3f(2) = remaindernofactor?

a = 2

f(2) = (2)³ − 3(2)² − (2) + 3 = −3Dividing gives x³ − 3x² − x + 3 = (x − 2)(x² − x − 3) − 3: the same remainder.The remainder is f(2), the height of the curve at x = 2. It is not 0, so (x − 2) is not a factor.

Slide $a$ and watch the remainder follow the height of the curve. Where the curve crosses the axis the remainder is 0, and $x - a$ is a factor.

Finding unknown coefficients

If a question tells you a factor or a remainder, substitute to get an equation. One unknown needs one fact; two unknowns need two facts, giving simultaneous equations.

Worked example · JAMB 2012

JAMB 2012 · UTME · Q12

Find the remainder when $2x^3 - 11x^2 + 8x - 1$ is divided by $x + 3$ .

Which value?
Dividing by $x + 3$ , so substitute $x = -3$ .
Think first. x + 3 = 0 when x = ?
Substitute
$2(-3)^3 = -54$ , $-11(-3)^2 = -99$ , $8(-3) = -24$ , and the constant is $-1$ .
Think first. Work out each term with x = −3.
Add
The remainder is $-178$ : option D.
Think first. −54 − 99 − 24 − 1 = ?

Worked example · JAMB 1994

JAMB 1994 · UME · Q12

Find the values of $p$ and $q$ such that $(x - 1)$ and $(x - 3)$ are factors of $px^3 + qx^2 + 11x - 6$ .

Use x − 1
$p + q + 11 - 6 = 0$ , so $p + q = -5$ .
Think first. f(1) = 0 gives which equation?
Use x − 3
$27p + 9q + 33 - 6 = 0$ , so $27p + 9q = -27$ . Divide by 9: $3p + q = -3$ .
Think first. f(3) = 0 gives which equation?
Solve together
$2p = 2$ , so $p = 1$ , and then $q = -6$ : option B.
Think first. Subtract the first equation from the second.

Your turn

JAMB 1995 · UME · Q11

Polynomials & algebraic division

If $x - 1$ and $x + 1$ are both factors of $x^3 + px^2 + qx + 6$ , evaluate $p$ and $q$ .

Worked solution (try it first)

By the factor theorem, the expression is 0 at $x = 1$ : $1 + p + q + 6 = 0$ , so $p + q = -7$ .
It is 0 at $x = -1$ : $-1 + p - q + 6 = 0$ , so $p - q = -5$ .
Add the equations: $2p = -12$ , so $p = -6$ and $q = -1$ , option A.

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