Find roots of polynomial $$ p(x) = 6x^4+23x^3+14x^2-53x-60 $$

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

$$ \begin{aligned}x_1 &= \frac{ 3 }{ 2 }\\[1 em]x_2 &= -\frac{ 4 }{ 3 }\\[1 em]x_3 &= -2+i\\[1 em]x_4 &= -2-i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 3 }{ 2 } } $ is a root of polynomial $ 6x^4+23x^3+14x^2-53x-60 $.

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}{ 60 } $, with factors of 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

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

The POSSIBLE zeroes are:

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

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( \dfrac{ 3 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 3 }{ 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}{ 2x-3 }$

$$ \frac{ 6x^4+23x^3+14x^2-53x-60}{ 2x-3} = 3x^3+16x^2+31x+20 $$

Step 2:

The next rational root is $ x = \dfrac{ 3 }{ 2 } $

$$ \frac{ 6x^4+23x^3+14x^2-53x-60}{ 2x-3} = 3x^3+16x^2+31x+20 $$

Step 3:

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

$$ \frac{ 3x^3+16x^2+31x+20}{ 3x+4} = x^2+4x+5 $$

Step 4:

The solutions of $ x^2+4x+5 = 0 $ are: $ x = -2+i ~ \text{and} ~ x = -2-i$.

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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