# 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,

$x - 2$)$2x^{3} + x^{2} - 7 x - 6$ ( $2 x^{2} + 5 x + 3$
$2 x^{3}-4 x^{2}$

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$5 x^{2}-7 x$
$5 x^{2}-10 x$

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$3 x-6$
$3 x-6$

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$0$

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$f(x)=(x-2)(2 x^{2}+5 x+3)$

$=(x-2)[2 x^{2}+2 x+3 x+3]$

$=(x-2)[2 x(x+1)+3(x+1)]$

$=(x-2)(x+1)(2 x+3)$

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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Updated on: 10-Oct-2022

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