Find roots of polynomial $$ p(x) = -6x^3-12x^2+78x+156 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= -2\\[1 em]x_2 &= \sqrt{ 13 }\\[1 em]x_3 &= -\sqrt{ 13 } \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = -2 } $ is a root of polynomial $ -6x^3-12x^2+78x+156 $.

The Rational Root Theorem tells us that if the polynomial has a rational zero then it must be a fraction $ \dfrac{ \color{blue}{p}}{ \color{red}{q} } $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.

The constant term is $ \color{blue}{ 156 } $, with factors of 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78 and 156.

The leading coefficient is $ \color{red}{ 6 }$, with factors of 1, 2, 3 and 6.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 156 }}{\text{ factors of 6 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156 ) }}{\text{ ( 1, 2, 3, 6 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 13}{ 1} \pm \frac{ 26}{ 1} \pm \frac{ 39}{ 1} \pm \frac{ 52}{ 1} \pm \frac{ 78}{ 1} \pm \frac{ 156}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 13}{ 2} \pm \frac{ 26}{ 2} \pm \frac{ 39}{ 2} \pm \frac{ 52}{ 2} \pm \frac{ 78}{ 2} \pm \frac{ 156}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 12}{ 3} \pm \frac{ 13}{ 3} \pm \frac{ 26}{ 3} \pm \frac{ 39}{ 3} \pm \frac{ 52}{ 3} \pm \frac{ 78}{ 3} \pm \frac{ 156}{ 3}\\[ 1 em] \pm \frac{ 1}{ 6} & \pm \frac{ 2}{ 6} & \pm \frac{ 3}{ 6} & \pm \frac{ 4}{ 6} & \pm \frac{ 6}{ 6} & \pm \frac{ 12}{ 6} & \pm \frac{ 13}{ 6} & \pm \frac{ 26}{ 6} & \pm \frac{ 39}{ 6} & \pm \frac{ 52}{ 6} & \pm \frac{ 78}{ 6} & \pm \frac{ 156}{ 6} ~~ \end{aligned} $$

Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

We can see that $ p\left( -2 \right) = 0 $ so $ x = -2 $ is a root of a polynomial $ p(x) $.

To find remaining zeros we use Factor Theorem. This theorem states that if $ \dfrac{p}{q} $ is root of the polynomial then the polynomial can be divided by $ \color{blue}{qx − p} $. In this example we divide polynomial $ p $ by $ \color{blue}{ x+2 }$

$$ \frac{ -6x^3-12x^2+78x+156}{ x+2} = -6x^2+78 $$

Step 2:

The next rational root is $ x = -2 $

$$ \frac{ -6x^3-12x^2+78x+156}{ x+2} = -6x^2+78 $$

Step 3:

The solutions of $ -6x^2+78 = 0 $ are: $ x = -\sqrt{ 13 } ~ \text{and} ~ x = \sqrt{ 13 }$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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