Determine the remainder when 9x5 4x4 is divided by 3x 1 Sh

Determine the remainder when 9x^5 - 4x^4 is divided by 3x - 1. Show, using the factor theorem, that 2x - 1 is a factor of 2x^4 - x^3 - 6x2 + 5x - 1 and hence express 2X^4 - x^3 - 6x^2 + 5x - 1 as a product of a linear and cubic factor.

Solution

4. (i) We have 9x⁵ – 4x⁴ = 3x⁴( 3x -1) + 3x⁴ – 4x⁴ = 3x⁴( 3x -1) – x⁴ = 3x⁴( 3x -1) – 1/3x³ ( 3x – 1) – 1.   Therefore, (9x⁵ – 4x⁴ )/ (3x – 1) = [ 3x⁴ – (1/3)x³ ] – 1/ ( 3x – 1). Thus, the remainder is - 1.

(ii) As per the factor theorem, (x – c) is a factor of the polynomial f (x) when f ( c) = 0. Thus, if 2x –1 is a factor of f(x) = 2x⁴– x³-6x²+ 5x – 1,then f (1/2) would be 0. Now, f (1/2) = 2( 1/16) – 1/8 – 6/4 +5/2 – 1 = 1/8 – 1/8 – 3/2 + 5/2 – 1 = 0. Therefore 2x – 1 is a factor of f (x) = 2x⁴– x³-6x²+ 5x – 1. Further, x⁴– x³- 6x²+ 5x – 1 = x³( 2x – 1) – 3x( 2x – 1)+ 1(2x – 1)= ( 2x -1)( x³– 3x + 1)