Question

use long division to find the quotient and the remainder when the function f(x) = x³+3x²-4 is divided by (x-1).hence factorize the function completely

Answer 1

$\begin{matrix} & x^2 +4x+4 \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ x-1 &)\overline{x^3+3x^2-4} \\ & - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \underline{x^3-x^2} \ \ \ \ \ \ \ \ \\ & \ \ \ \ \ \ 4x^2-4 \\ & \ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \underline{4x^2-4x} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4x-4 \\ & \ \ - \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{4x-4} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ \end{matrix}$

Therefore,

$$\dfrac{x^3+3x^2-4}{x-1}=x^2+4x+4=(x+2)^2$$

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