Question

Divide:
(6+x-4x^2+x^3)by(x-3

Answer 1

this is answer of question

Answer 2

Answer:

x^2-x-2

Step-by-step explanation:

as per the question:

$\left(\begin{array}{ccc}\underline{6+x-4x^2+x^3}\\x-3\end{array}\right)\\=\left(\begin{array}{ccc}\underline{x^3-4x^2+x+6}\\x-3\end{array}\right)$

[rearranging]

$In~~\left(\begin{array}{ccc}\underline{x^3-4x^2+x+6}\\x-3\end{array}\right),factorize ~x^3-4x^2+x+6$

this gives:

$x^3-4x^2+x+6\\=x^3+x^2-5x^2-5x+6x+6\\=x^2(x+1)-5x(x+1)+6(x+1)\\=(x^2-5x+6)(x+1)\\=(x^2-3x-2x+6)(x+1)\\=[x(x-3)-2(x-3)](x+1)\\=(x-2)(x-3)(x+1)$

on substitution:

$\left(\begin{array}{ccc}\underline{(x-2)(x-3)(x+1)}\\x-3\end{array}\right)\\=(x-2)(x+1)\\=x^2-x-2$