Evaluation

Video lesson · from 18:311 worked example

One lesson video covers all of Polynomials & Quadratics, so it opens at 18:31 for this topic — not from the beginning.

Theory

Polynomials can be evaluated using substitution.

For example, if we take $f(x)=2x^3-9x^2+2x+1$ and we evaluate $f(5)$ , we obtain:

f(5) = 2(5)^3 - 9(5)^2 + 2(5) + 1 = 36

This means that if we divide $f(x)$ by $(x-5)$ , we have a remainder of 36.

If there is no remainder when you evaluate $f(a)$ , i.e. $f(a)=0$ , this means that $x=a$ is a root of $f(x)$ and therefore $(x-a)$ is a factor of $f(x)$ .

For $g(x)=x^3+x^2-22x-40$ , evaluating gives $g(-2)=0$ .

This means that $x=-2$ is a root of $g(x)$ and $(x+2)$ is a factor of $g(x)$ .

We can use this to help with factorisation:

g(x) = (x+2)q(x)

where $q(x)$ is a quadratic in this case.

Task: Complete the table

Root	Associated Factor
$x = -2$	$x+2$
$x = 5$	$x-5$
$x = -7$	$x+7$
$x = 7$	$x-7$
$x = 3$	$x-3$
$x = -3$	$x+3$
$x = \frac{1}{2}$	$2x-1$
$x = \frac{1}{3}$	$3x-1$
$x = \frac{2}{3}$	$3x-2$
$x = -\frac{5}{4}$	$4x+5$

⚠️ Common Examiner Traps

Include zeros for missing powers: for $x^3 - 7x + 6$ the row is 1, 0, −7, 6. Omitting the zero is the most common synthetic division error there is.
Divide by $a$ , not $-a$ : to test the factor $(x-3)$ you use 3 in the box. The sign reverses.
Set it out in the standard layout: non-standard arrangements are hard to credit, even when the arithmetic is right.
The last number is the remainder: if it is zero, say explicitly that this proves the divisor is a factor.

Worked examples

Example 1

Given $g(x)=x^3+x^2-22x-40$ , evaluate the following and state the remainder:

g(3)

g(-5)

g(-2)

a) $g(3)$

g(3) = (3)^3 + (3)^2 - 22(3) - 40

g(3) = 27 + 9 - 66 - 40 = -70

The remainder is $-70$ .

b) $g(-5)$

g(-5) = (-5)^3 + (-5)^2 - 22(-5) - 40

g(-5) = -125 + 25 + 110 - 40 = -30

The remainder is $-30$ .

c) $g(-2)$

g(-2) = (-2)^3 + (-2)^2 - 22(-2) - 40

g(-2) = -8 + 4 + 44 - 40 = 0

The remainder is $0$ .