2010 Paper 1 Q1 - Remainder & Factor Theorem

Let’s look at the first question of the 2010 paper 1…

1. The function f is defined by  f(x) = x⁴ – x³ + kx -4, where k is a constant.

(i) given that x – 2 is a factor of f(x), find the value of k.

(ii) Using the value of k found in part (i), find the remainder when f(x) is divided by x + 2.

We must remember that, if x – 2 is a factor, then f(2) = 0. And this is the way we solve for k.

f(2) = 2⁴ – 2³ + 2k – 4 = 0

Therefore,  16 – 8 + 2k – 4 = 0

                                4 + 2k = 0

                                2k = -4

                                k = -2

Once we have found the value of k, it would be useful to end this part of the question by putting it into the f(x) expression.

Hence,

                f(x) = x⁴ – x³ – 2x – 4

(ii)  To find the remainder when f(x) is divided by  x + 2, we find the value of f(-2).  Why f(-2) and not f(2)? Well, if we equate the divisor, x + 2, to zero, we will then be able to solve for x which is -2.  Hence, this is the value which we substitute for x in the expression for f(x).

So,

                f(-2) = (-2)⁴ – (-2)³ -2(-2) – 4

                        = 16 –(-8) + 4 – 4

                        = 24

Hence, the remainder when f(x) is divided by x + 2 is 24.

This is really a slam dunk question which, hopefully, everyone got correct for the O levels.

Check out the answers to the rest of the questions…..