Question

Prove by factor theorem that,
i) $(x-2)\text{is a factor of}2x^3-7x–2$.

Answer 1

To prove: $(x-2)\text{is a factor of}2x^3-7x–2$.
From Factor theorem, we know that a polynomial $f\left(x\right)$ has a factor $(x–a)$ if and only if $f\left(a\right)=0$.
Here, $a=2$
$f\left(a\right)=f\left(2\right)$
$=2(2{)}^3-7(2)-2$
$=2\left(8\right)-7\left(2\right)-2$
$=16-14-2$
$=16-16$
$=0$
$\therefore f\left(a\right)=0$
Hence, $(x-2)$ is a factor of $2x^3-7x-2$.