Find roots of polynomial $$ p(x) = 5x^4-28x^3+33x^2-2x-8 $$

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

$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 4\\[1 em]x_3 &= -\frac{ 2 }{ 5 }\\[1 em]x_4 &= 1 \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 5x^4-28x^3+33x^2-2x-8 $.

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

The leading coefficient is $ \color{red}{ 5 }$, with factors of 1 and 5.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 8 }}{\text{ factors of 5 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8 ) }}{\text{ ( 1, 5 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 4}{ 5} \pm \frac{ 8}{ 5} ~~ \end{aligned} $$

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( 1 \right) = 0 $ so $ x = 1 $ 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-1 }$

$$ \frac{ 5x^4-28x^3+33x^2-2x-8}{ x-1} = 5x^3-23x^2+10x+8 $$

Step 2:

The next rational root is $ x = 1 $

$$ \frac{ 5x^4-28x^3+33x^2-2x-8}{ x-1} = 5x^3-23x^2+10x+8 $$

Step 3:

The next rational root is $ x = 4 $

$$ \frac{ 5x^3-23x^2+10x+8}{ x-4} = 5x^2-3x-2 $$

Step 4:

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

$$ \frac{ 5x^2-3x-2}{ 5x+2} = x-1 $$

Step 5:

To find the last zero, solve equation $ x-1 = 0 $

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

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