Question

Find the rational roots of the polynomial $f\left(x\right)=2x^2+x^2-7x-6$.

Answer 1

Given: $f\left(x\right)=2x^2+x^2-7x-6$

The rational roots of the $f\left(x\right)$ is limited to the factors of the product of leading coefficient and constant.

$\because$ Factors of $-12=\pm 1,\pm 2,\pm 3$
$\pm \frac{1}{2},\pm \frac{3}{2}$ and $\pm 6$

We observe that,

$f(-1)=2\times (-1{)}^3+(-1{)}^2-7\times (-1)-6$

$=-2+1+7-6=0$

$f\left(2\right)=2\times (2{)}^3+(2{)}^2-7\times \left(2\right)-6$

$=16+4-14-6=0$

And $f(-\frac{3}{2})=2\times {(-\frac{3}{2})}^3+{(-\frac{3}{2})}^2-7\times (-\frac{3}{2})-6$

$=2\times -\frac{27}{8}+\frac{9}{4}+\frac{21}{2}-6=0$

$\because f(-1)=0,f(-2)$ and $f(-\frac{3}{2})=0$

Hence, $-1,-2$ and $-\frac{3}{2}$ are the integral roots of the polynomial $f\left(x\right)=2x^3+x^2-7x-6$.