f(x) = 2x^3 - 3x^2 - 39x + 20 (a) Use the factor theorem to show that (x + 4) is a factor of f(x) - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 2

To factorise f(x) completely, we already know one factor is (x + 4). We can perform polynomial long division or synthetic division to find the other factor:

Dividing $f(x)$ by $(x + 4)$, we get:

1. Perform the division: $f(x) = (x + 4)(2x^2 - 11x + 5)$

2. Now, factor the quadratic $2x^2 - 11x + 5$:

   • Finding its roots using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = -11$, and $c = 5$: $x = \frac{11 \pm \sqrt{(-11)^2 - 4(2)(5)}}{2(2)}$ $= \frac{11 \pm \sqrt{121 - 40}}{4}$ $= \frac{11 \pm \sqrt{81}}{4}$ $= \frac{11 \pm 9}{4}$ The roots are: $x = \frac{20}{4} = 5$ and $x = \frac{2}{4} = \frac{1}{2}$

Thus, we can write: $2x^2 - 11x + 5 = 2(x - 5)(x - \frac{1}{2})$

Hence, the complete factorization of f(x) is: $f(x) = (x + 4)(2(x - 5)(x - \frac{1}{2}))$

This gives us: $f(x) = 2(x + 4)(x - 5)(2x - 1)$