Find roots of polynomial $$ p(x) = 6x^5+31x^4+23x^3-20x^2 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -4\\[1 em]x_3 &= \frac{ 1 }{ 2 }\\[1 em]x_4 &= -\frac{ 5 }{ 3 } \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x^2 }$ from $ 6x^5+31x^4+23x^3-20x^2 $ and solve two separate equations:

$$ \begin{aligned} 6x^5+31x^4+23x^3-20x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( 6x^3+31x^2+23x-20 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ 6x^3+31x^2+23x-20 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

Use rational root test to find out that the $ \color{blue}{ x = -4 } $ is a root of polynomial $ 6x^3+31x^2+23x-20 $.

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}{ 20 } $, with factors of 1, 2, 4, 5, 10 and 20.

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

The POSSIBLE zeroes are:

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

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( -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 }$

$$ \frac{ 6x^3+31x^2+23x-20}{ x+4} = 6x^2+7x-5 $$

Step 3:

The next rational root is $ x = -4 $

$$ \frac{ 6x^3+31x^2+23x-20}{ x+4} = 6x^2+7x-5 $$

Step 4:

The next rational root is $ x = \dfrac{ 1 }{ 2 } $

$$ \frac{ 6x^2+7x-5}{ 2x-1} = 3x+5 $$

Step 5:

To find the last zero, solve equation $ 3x+5 = 0 $

$$ \begin{aligned} 3x+5 & = 0 \\[1 em] 3 \cdot x & = -5 \\[1 em] x & = - \frac{ 5 }{ 3 } \end{aligned} $$

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