### A. REMAINDER THEOREM

If ** f(x)** is divided by (x

**–**a) ⇒ the remainder is

*f(a)*Equate (x-a) to zero all the time

x-a = 0 –> x=a –> substitute “a” into x

Some of you may be irritated by the f and (x) and a, here’s a easier way to understand … go straight to the Polynomial Questions

#### Remainder Theorem Question:

Find the remainder when 5x^{3} – 5x + 1 is divided by:

i. x-2, ii. x+3, iii. 2x-1

Ans: Let ** f(x) **= 4x

^{3}– 5x + 1

f(2) = 5(2)^{3}-5(2)+1 = 31 = Remainder

f(-3) = 5(-3)^{3}-5(-3)+1 = -119 = Remainder

it’s divided by x

+3 so since x+3=0 –> x=-3 –> substitute “-3” into x

f(½) = 5(½)^{3}-5(½)+1 = -0.875

it’s divided by (2x-1) so since 2x-1 = 0 –> x=1/2 –>

substitute “1/2” into x

DON’T waste time doing long division in remainder theorem questions!!!

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