1. Lesson 4-Remainder Theorem 17 December, 2015ML4 MH Objectives : - The remainder and Factor theorems - It’s used to help factorise Polynomials - It’s used to help find roots of equation - It’s used to help find remainder without actually dividing

2. f(x) Notation f(x) stands for “factor of x” f(x) notation is often used instead of “y” So we can say 17 December, 2015ML4 MH f(0) = 0 3 -7(0) 2 +5(0)-6 = -6 f(1) = 1 3 -7(1) 2 +5(1)-6 = -7 f(2) = 2 3 -7(2) 2 +5(2)-6 = -16 f(-1)= (-1) 3 -7(-1) 2 +5(-1)-6 = -19

3. Consider this division __x2-x-12____ (x-2) | x 3 -3x 2 -10x+24 x 3 -2x 2 -x2-10x -x2+ 2x - 12x+24 0 17 December, 2015ML4 MH x-2 is a factor of x 3 -3x 2 -10x+24 Calculate f(2) f(2)=0 If f(a)=0, then (x-a) is a factor of the Polynomial so if f(a) = 0 the remainder is zero

4. Consider this division 17 December, 2015ML4 MH Calculate f(-2) What do you notice f(-2)= 2(-2) 3 +3(-2) 2- (-2)+1 f(-2)=-1 If a polynomial f(x) is divided by (x – a), the remainder is the constant f(a), and f(x) = q(x) ∙ (x – a) + f(a) where q(x) is a polynomial with degree one less than the degree of f(x). Why is this ??

5. This is because 17 December, 2015ML4 MH A division is essentially : Divisor d(x) is of the form (x-a) When x=a i.e. f(a) if f(a)=0 -> factor (x-a) otherwise f(a)=R -> (x-a) gives R Do exercise 3c page 94

6. Summary If a polynomial f(x) is divided by (x – a), the remainder is the constant f(a), and f(x) = q(x) ∙ (x – a) + f(a) where q(x) is a polynomial with degree one less than the degree of f(x). 17 December, 2015ML4 MH