Question

Use remainder theorem to factorize: 2x^3+3x^2-9x-10

Answer 1

Answer:

$\textsf{The factors are (x-2), (x+1) and (2x+5)}$

Step-by-step explanation:

$\text{Let}\bf\,P(x)=2x^3+3x^2-9x-10$

$\text{put x=2}$

$P(2)=2(2)^3+3(2)^2-9(2)-10$

$P(2)=16+12-18-10$

$P(2)=0$

$\therefore\textbf{(x-2) is a factor of P(x)}$

$\text{put x=-1}$

$P(-1)=2(-1)^3+3(-1)^2-9(-1)-10$

$P(-1)=-2+3+9-10$

$P(-1)=0$

$\therefore\textbf{(x+1) is a factor of P(x)}$

$\text{By synthetic division, we have}$

$\begin{array}{r|rrrr}2&2&3&-9&-10\\&&4&14&10\\\cline{2-5}\\-1&2&7&5&|\;0\\&&-2&-5&\\\cline{2-5}\\&2&5&|\;0&\end{array}$

$\therefore\textbf{The remaining factor is (2x+5)}$

Answer 2

Step-by-step explanation:

refer to the answer above