Find roots of polynomial $$ p(x) = x^5+2x^4-30x^3+20x^2+325x-750 $$

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

$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -5\\[1 em]x_3 &= -5\\[1 em]x_4 &= \frac{ 5 }{ 2 }+\frac{\sqrt{ 15 }}{ 2 }i\\[1 em]x_5 &= \frac{ 5 }{ 2 }- \frac{\sqrt{ 15 }}{ 2 }i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ x^5+2x^4-30x^3+20x^2+325x-750 $.

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}{ 750 } $, with a single factor of 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 125, 150, 250, 375 and 750.

The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 750 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 125, 150, 250, 375, 750 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 75}{ 1} \pm \frac{ 125}{ 1} \pm \frac{ 150}{ 1} \pm \frac{ 250}{ 1} \pm \frac{ 375}{ 1} \pm \frac{ 750}{ 1} ~~ \end{aligned} $$

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

$$ \frac{ x^5+2x^4-30x^3+20x^2+325x-750}{ x-3} = x^4+5x^3-15x^2-25x+250 $$

Step 2:

The next rational root is $ x = 3 $

$$ \frac{ x^5+2x^4-30x^3+20x^2+325x-750}{ x-3} = x^4+5x^3-15x^2-25x+250 $$

Step 3:

The next rational root is $ x = -5 $

$$ \frac{ x^4+5x^3-15x^2-25x+250}{ x+5} = x^3-15x+50 $$

Step 4:

The next rational root is $ x = -5 $

$$ \frac{ x^3-15x+50}{ x+5} = x^2-5x+10 $$

Step 5:

The solutions of $ x^2-5x+10 = 0 $ are: $ x = \dfrac{ 5 }{ 2 }+\dfrac{\sqrt{ 15 }}{ 2 }i ~ \text{and} ~ x = \dfrac{ 5 }{ 2 }-\dfrac{\sqrt{ 15 }}{ 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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