Question

Factorise x^4+x^3+2x^2+4x-8 using factor theorem

Answer 1

Answer:

(x-1)(x+2)(x²+4)

Step-by-step explanation:

$let, f(x)= x^4+x^3+2x^2+4x-8$

$here, f(1)= 1^4+1^3+2\times1^2+4\times1-8$

               $= 1+1+2+4-8$

               $= 8-8$

               $=0$

$so,\ x-1 \ \ is \ a \ factor \ of \ f(x)$

$now, f(x)= x^4+x^3+2x^2+4x-8$

$= x^4-x^3+2x^3-2x^2+4x^2-4x+8x-8$

$= x^3(x-1)+2x^2(x-1)+4x(x-1)+8(x-1)$

$=(x-1)(x^3+2x^2+4x+8)$

$=(x-1)(x^2(x+2)+4(x+2))$

$=(x-1)(x+2)(x^2+4)$