Find roots of polynomial $$ p(x) = r^5+5r^4-7r^3-43r^2-8r-48 $$

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

$$ \begin{aligned}r_1 &= 3\\[1 em]r_2 &= -4\\[1 em]r_3 &= -4\\[1 em]r_4 &= i\\[1 em]r_5 &= -i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ r = 3 } $ is a root of polynomial $ r^5+5r^4-7r^3-43r^2-8r-48 $.

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}{ 48 } $, with a single factor of 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

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

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 48 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 48}{ 1} ~~ \end{aligned} $$

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}{ r-3 }$

$$ \frac{ r^5+5r^4-7r^3-43r^2-8r-48}{ r-3} = r^4+8r^3+17r^2+8r+16 $$

Step 2:

The next rational root is $ r = 3 $

$$ \frac{ r^5+5r^4-7r^3-43r^2-8r-48}{ r-3} = r^4+8r^3+17r^2+8r+16 $$

Step 3:

The next rational root is $ r = -4 $

$$ \frac{ r^4+8r^3+17r^2+8r+16}{ r+4} = r^3+4r^2+r+4 $$

Step 4:

The next rational root is $ r = -4 $

$$ \frac{ r^3+4r^2+r+4}{ r+4} = r^2+1 $$

Step 5:

The solutions of $ r^2+1 = 0 $ are: $ r = i ~ \text{and} ~ r = -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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