Find roots of polynomial $$ p(x) = x^5-3x^3+8x^2-24 $$

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

$$ \begin{aligned}x_1 &= -2\\[1 em]x_2 &= 1.7321\\[1 em]x_3 &= -1.7321\\[1 em]x_4 &= 1+1.7321i\\[1 em]x_5 &= 1-1.7321i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = -2 } $ is a root of polynomial $ x^5-3x^3+8x^2-24 $.

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}{ 24 } $, with a single factor of 1, 2, 3, 4, 6, 8, 12 and 24.

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

The POSSIBLE zeroes are:

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

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( -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 }$

$$ \frac{ x^5-3x^3+8x^2-24}{ x+2} = x^4-2x^3+x^2+6x-12 $$

Step 2:

The next rational root is $ x = -2 $

$$ \frac{ x^5-3x^3+8x^2-24}{ x+2} = x^4-2x^3+x^2+6x-12 $$

Step 3:

Polynomial $ x^4-2x^3+x^2+6x-12 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.

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