Question

Prove by factor theorem that,
(iv) $(2x-1)\text{is a factor of}6x^3-x^2-5x+2$.

Answer 1

To prove: $(2x-1)\text{is a factor of}6x^3-x^2-5x+2$.
From Factor theorem, we know that A polynomial $f\left(x\right)$ has a factor $(x–a)$ if and only if $f\left(a\right)=0$.
$2x-1=0\Rightarrow x=\frac{1}{2}$
So $a=\frac{1}{2}$,
$f\left(a\right)=f\left(\frac{1}{2}\right)$
$=6{\left(\frac{1}{2}\right)}^3-{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+2$
$=6\left(\frac{1}{8}\right)-\left(\frac{1}{4}\right)-\left(\frac{5}{2}\right)+2$
$=\left(\frac{3}{4}\right)-\left(\frac{1}{4}\right)-\left(\frac{5}{2}\right)+2$
$=\left(\frac{1}{2}\right)-\left(\frac{5}{2}\right)+2$
$=-\left(\frac{4}{2}\right)+2$
$=-2+2$
$=0$
Hence,$(2x-1)\text{is a factor of}6x^3-x^2-5x+2$.