Find roots of polynomial $$ p(x) = -120x^3+124x^2-3x-1 $$

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

$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 0.1095\\[1 em]x_3 &= -0.0761 \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ -120x^3+124x^2-3x-1 $.

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}{ 1 } $, with factors of 1.

The leading coefficient is $ \color{red}{ 120 }$, with factors of 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 1 }}{\text{ factors of 120 }} = \pm \dfrac{\text{ ( 1 ) }}{\text{ ( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1}\\[ 1 em] \pm \frac{ 1}{ 2}\\[ 1 em] \pm \frac{ 1}{ 3}\\[ 1 em] \pm \frac{ 1}{ 4}\\[ 1 em] \pm \frac{ 1}{ 5}\\[ 1 em] \pm \frac{ 1}{ 6}\\[ 1 em] \pm \frac{ 1}{ 8}\\[ 1 em] \pm \frac{ 1}{ 10}\\[ 1 em] \pm \frac{ 1}{ 12}\\[ 1 em] \pm \frac{ 1}{ 15}\\[ 1 em] \pm \frac{ 1}{ 20}\\[ 1 em] \pm \frac{ 1}{ 24}\\[ 1 em] \pm \frac{ 1}{ 30}\\[ 1 em] \pm \frac{ 1}{ 40}\\[ 1 em] \pm \frac{ 1}{ 60}\\[ 1 em] \pm \frac{ 1}{ 120}\\[ 1 em] \end{aligned} $$

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( 1 \right) = 0 $ so $ x = 1 $ 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-1 }$

$$ \frac{ -120x^3+124x^2-3x-1}{ x-1} = -120x^2+4x+1 $$

Step 2:

The next rational root is $ x = 1 $

$$ \frac{ -120x^3+124x^2-3x-1}{ x-1} = -120x^2+4x+1 $$

Step 3:

The solutions of $ -120x^2+4x+1 = 0 $ are: $ x = \dfrac{ 1 }{ 60 }-\dfrac{\sqrt{ 31 }}{ 60 } ~ \text{and} ~ x = \dfrac{ 1 }{ 60 }+\dfrac{\sqrt{ 31 }}{ 60 }$.

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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