Find the rational roots of the polynomial f

Solution:

Given that $f(x)=2 x^{3}+x^{2}-7 x-6$

f(x) is a cubic polynomial with an integer coefficient . If the rational root in the form of p/q, the values of p are limited to factors of 6 which are ± 1, ± 2, ± 3, ± 6

and the values of q are limited to the highest degree coefficient i.e 2 which are ±1, ±2

here, the possible rational roots are ± 1, ± 2, ± 3, ± 6, ± 1/2, ± 3/2

Let, x = -1

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

= - 2 + 1 + 7 - 6

= - 8 + 8

= 0

Let, x = 2

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

= (2 * 8) + 4 - 14 - 6

= 16 + 4 -14 - 6

= 20 - 20

= 0

Let, x = -3/2

$f(-3 / 2)=2(-3 / 2)^{3}+(-3 / 2)^{2}-7(-3 / 2)-6$

= 2(-27/8) + 9/4 - 7(-3/2) - 6

= (−27/4) + 9/4 - (-21/2) - 6

= – 6.75 + 2.25 + 10.5 - 6

= 12.75 - 12.75

= 0

But from all the factors only -1, 2 and −3/2 gives the result as zero

So, the rational roots of $2 x^{3}+x^{2}-7 x-6$ are $-1,2$ and $-3 / 2$