Find roots of polynomial $$ p(x) = -2x^4+9x^3+30x^2-52x-120 $$

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

$$ \begin{aligned}x_1 &= 6\\[1 em]x_2 &= -2\\[1 em]x_3 &= \frac{ 5 }{ 2 }\\[1 em]x_4 &= -2 \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 6 } $ is a root of polynomial $ -2x^4+9x^3+30x^2-52x-120 $.

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}{ 120 } $, with factors of 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.

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

The POSSIBLE zeroes are:

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

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( 6 \right) = 0 $ so $ x = 6 $ 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-6 }$

$$ \frac{ -2x^4+9x^3+30x^2-52x-120}{ x-6} = -2x^3-3x^2+12x+20 $$

Step 2:

The next rational root is $ x = 6 $

$$ \frac{ -2x^4+9x^3+30x^2-52x-120}{ x-6} = -2x^3-3x^2+12x+20 $$

Step 3:

The next rational root is $ x = -2 $

$$ \frac{ -2x^3-3x^2+12x+20}{ x+2} = -2x^2+x+10 $$

Step 4:

The next rational root is $ x = \dfrac{ 5 }{ 2 } $

$$ \frac{ -2x^2+x+10}{ 2x-5} = -x-2 $$

Step 5:

To find the last zero, solve equation $ -x-2 = 0 $

$$ \begin{aligned} -x-2 & = 0 \\[1 em] -1 \cdot x & = 2 \\[1 em] x & = \frac{ 2 }{ -1 } \end{aligned} $$

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