Question

Show that $(x-2)$ is a factor of the polynomial $f\left(x\right)=2x^3-3x^2-17x+30$ and hence factorize f(x)

Answer 1

Let $(x-2)$ be a factor

$\therefore x=2$ multify

$f\left(x\right)=0$

$\therefore 2x^3-3x^2-17x+30=0$ at $x=2$

$\therefore 2(2{)}^3-3(2{)}^2-17\left(2\right)+30=0$

$\therefore 16-12-34+30=0$

$\therefore 4-4=0$

$\therefore 0=0$

Now, $x-2|\stackrel{2x^2+x-15}{\overline{2x^3-3x^2-17x+30}}$

$\underset{–}{\underset{-}{2}{x}^3\underset{+}{-}4x^2}$

$x^2-17x$

$\underset{–}{\underset{-}{x^2}\underset{+}{-}2x}$

$-15x+30$

$\underset{–}{\underset{+}{-}15x\underset{-}{+}30}$

$0$

$\therefore 2x^3-3x^2-17x+30=(x-2)(2x^2+x-15)$

$=(x-2)(2x-5)(x+3)$

Ans :- $(x-2),(2x-5),(x+3)$ are factors.