Find roots of polynomial $$ p(x) = x^5-2x^4-6x^3+4x^2+13x+6 $$

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

$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= 3\\[1 em]x_3 &= -1\\[1 em]x_4 &= -1\\[1 em]x_5 &= -1 \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ x^5-2x^4-6x^3+4x^2+13x+6 $.

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}{ 6 } $, with a single factor of 1, 2, 3 and 6.

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

The POSSIBLE zeroes are:

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

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 }$

$$ \frac{ x^5-2x^4-6x^3+4x^2+13x+6}{ x-2} = x^4-6x^2-8x-3 $$

Step 2:

The next rational root is $ x = 2 $

$$ \frac{ x^5-2x^4-6x^3+4x^2+13x+6}{ x-2} = x^4-6x^2-8x-3 $$

Step 3:

The next rational root is $ x = 3 $

$$ \frac{ x^4-6x^2-8x-3}{ x-3} = x^3+3x^2+3x+1 $$

Step 4:

The next rational root is $ x = -1 $

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

Step 5:

The next rational root is $ x = -1 $

$$ \frac{ x^2+2x+1}{ x+1} = x+1 $$

Step 6:

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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