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Find the rational roots of the polynomial $f(x) = 2x^3 + x^2 - 7x - 6$.
Given :
The given polynomial is $f(x) = 2x^3 + x^2 - 7x - 6$.
To find :
We have to find the rational roots of the polynomial $f(x) = 2x^3 + x^2 - 7x - 6$.
Solution :
$f(x) = 2x^3 + x^2 - 7x - 6$.
Here, $f(x)$ is a polynomial with integer coefficient and the coefficient of the highest degree term is 2.
Therefore, integer roots of $f(x)$ are limited to the integer factors of $2\times-6=-12$, which are $\pm 1, \pm 2, \pm 3, \pm 6$
If $x=1$,
$f(1)=2(1)^{3}+(1)^{2}-7 \times 1-6$
$=2 \times 1+1-7-6$
$=2+1-7-6$
$=3-13$
$=-10$
Therefore, $x=1$ is not a zero of $f(x)$.
Similarly,
$f(2)=2 \times(2)^{3}+(2)^{2}-7 \times 2-6$
$=2 \times 8+4-14-6$
$=16+4-14-6$
$=20-20$
$=0$
Therefore, $x=2$ is a zero of $f(x)$.
$f(3)=2(3)^{3}+(3)^{2}-7 \times 3-6$
$=2 \times 27+9-21-6$
$=54+9-21-6$
$=63-27$
$=36$
Therefore, $x=3$ is not a zero of $f(x)$.
$f(-3)=2(-3)^{3}+(-3)^{2}-7(-3)-6$
$=2 \times(-27)+9+21-6$
$=-54+9+21-6$
$=-60+30$
$=-30$
Therefore, $x=-3$ is not a zero of $f(x)$.
Dividing $f(x)$ by $x-2$, we get,
$x - 2$)$ 2x^{3} + x^{2} - 7 x - 6$ ( $2 x^{2} + 5 x + 3$
$2 x^{3}-4 x^{2}$
--------------------------
$5 x^{2}-7 x$
$5 x^{2}-10 x$
------------------------
$3 x-6$
$3 x-6$
-------------
$0$
--------------
$f(x)=(x-2)(2 x^{2}+5 x+3)$
$=(x-2)[2 x^{2}+2 x+3 x+3]$
$=(x-2)[2 x(x+1)+3(x+1)]$
$=(x-2)(x+1)(2 x+3)$
If $x-2=0$, $\Rightarrow x=2$
If $x+1=0$, $\Rightarrow x=-1$
If $x+3=0$, $\Rightarrow 2 x=-3$
$\Rightarrow x=\frac{-3}{2}$
The rational roots of the polynomial $f(x) = 2x^3 + x^2 - 7x - 6$ are $-1, 2, \frac{-3}{2}$.
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