WAEC 2011 · Paper 2 · Q1

(a)
If $f(x) = 6x^3 + 13x^2 + 2x - 5$ and $f(-1) = 0$ , find the factors of $f(x)$ .

Worked solution (try it first)

$f(-1) = 0$ , so by the factor theorem $(x + 1)$ is a factor.
Divide by $x + 1$ with synthetic division: write $-1$ on the left and $6, 13, 2, -5$ .
Bring down 6.
Then $6 \times (-1) = -6$ and $13 - 6 = 7$ .
$7 \times (-1) = -7$ and $2 - 7 = -5$ .
Last, $-5 \times (-1) = 5$ and $-5 + 5 = 0$ : no remainder.
So $f(x) = (x + 1)(6x^2 + 7x - 5)$ .
Factorise the quadratic: two numbers that multiply to $6 \times (-5) = -30$ and add to 7 are 10 and $-3$ .
So $6x^2 + 10x - 3x - 5 = 2x(3x + 5) - 1(3x + 5)$
$= (2x - 1)(3x + 5)$ .
The factors are $(x + 1)$ , $(2x - 1)$ and $(3x + 5)$ : $f(x) = (x + 1)(2x - 1)(3x + 5)$ .