Find roots of polynomial $$ p(x) = 12x^3+49x^2-16x-80 $$

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

$$ \begin{aligned}x_1 &= -4\\[1 em]x_2 &= -\frac{ 4 }{ 3 }\\[1 em]x_3 &= \frac{ 5 }{ 4 } \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = -4 } $ is a root of polynomial $ 12x^3+49x^2-16x-80 $.

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}{ 80 } $, with factors of 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.

The leading coefficient is $ \color{red}{ 12 }$, with factors of 1, 2, 3, 4, 6 and 12.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 80 }}{\text{ factors of 12 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 8, 10, 16, 20, 40, 80 ) }}{\text{ ( 1, 2, 3, 4, 6, 12 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 40}{ 1} \pm \frac{ 80}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 16}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 40}{ 2} \pm \frac{ 80}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 10}{ 3} \pm \frac{ 16}{ 3} \pm \frac{ 20}{ 3} \pm \frac{ 40}{ 3} \pm \frac{ 80}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 16}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 40}{ 4} \pm \frac{ 80}{ 4} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 4}{ 6} \pm \frac{ 5}{ 6} \pm \frac{ 8}{ 6} \pm \frac{ 10}{ 6} \pm \frac{ 16}{ 6} \pm \frac{ 20}{ 6} \pm \frac{ 40}{ 6} \pm \frac{ 80}{ 6} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 2}{ 12} \pm \frac{ 4}{ 12} \pm \frac{ 5}{ 12} \pm \frac{ 8}{ 12} \pm \frac{ 10}{ 12} \pm \frac{ 16}{ 12} \pm \frac{ 20}{ 12} \pm \frac{ 40}{ 12} \pm \frac{ 80}{ 12} ~~ \end{aligned} $$

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( -4 \right) = 0 $ so $ x = -4 $ 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+4 }$

$$ \frac{ 12x^3+49x^2-16x-80}{ x+4} = 12x^2+x-20 $$

Step 2:

The next rational root is $ x = -4 $

$$ \frac{ 12x^3+49x^2-16x-80}{ x+4} = 12x^2+x-20 $$

Step 3:

The next rational root is $ x = -\dfrac{ 4 }{ 3 } $

$$ \frac{ 12x^2+x-20}{ 3x+4} = 4x-5 $$

Step 4:

To find the last zero, solve equation $ 4x-5 = 0 $

$$ \begin{aligned} 4x-5 & = 0 \\[1 em] 4 \cdot x & = 5 \\[1 em] x & = \frac{ 5 }{ 4 } \end{aligned} $$

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