Find roots of polynomial $$ p(x) = 2x^5+7x^4+4x^3-36x^2-48x+32 $$

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

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

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 2x^5+7x^4+4x^3-36x^2-48x+32 $.

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}{ 32 } $, with factors of 1, 2, 4, 8, 16 and 32.

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

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 32 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8, 16, 32 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 32}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 16}{ 2} \pm \frac{ 32}{ 2} ~~ \end{aligned} $$

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

$$ \frac{ 2x^5+7x^4+4x^3-36x^2-48x+32}{ x-2} = 2x^4+11x^3+26x^2+16x-16 $$

Step 2:

The next rational root is $ x = 2 $

$$ \frac{ 2x^5+7x^4+4x^3-36x^2-48x+32}{ x-2} = 2x^4+11x^3+26x^2+16x-16 $$

Step 3:

The next rational root is $ x = -2 $

$$ \frac{ 2x^4+11x^3+26x^2+16x-16}{ x+2} = 2x^3+7x^2+12x-8 $$

Step 4:

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

$$ \frac{ 2x^3+7x^2+12x-8}{ 2x-1} = x^2+4x+8 $$

Step 5:

The solutions of $ x^2+4x+8 = 0 $ are: $ x = -2+2i ~ \text{and} ~ x = -2-2i$.

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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