Find roots of polynomial $$ p(x) = x^6-2x^5-11x^4+12x^3+36x^2 $$

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= 3\\[1 em]x_3 &= -2\\[1 em]x_4 &= 3\\[1 em]x_5 &= -2 \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x^2 }$ from $ x^6-2x^5-11x^4+12x^3+36x^2 $ and solve two separate equations:

$$ \begin{aligned} x^6-2x^5-11x^4+12x^3+36x^2 & = 0\\[1 em] \color{blue}{ x^2 }\cdot ( x^4-2x^3-11x^2+12x+36 ) & = 0 \\[1 em] \color{blue}{ x^2 = 0} ~~ \text{or} ~~ x^4-2x^3-11x^2+12x+36 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ x^4-2x^3-11x^2+12x+36 $.

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}{ 36 } $, with a single factor of 1, 2, 3, 4, 6, 9, 12, 18 and 36.

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

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 36 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 9, 12, 18, 36 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 36}{ 1} ~~ \end{aligned} $$

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}{ x-3 }$

$$ \frac{ x^4-2x^3-11x^2+12x+36}{ x-3} = x^3+x^2-8x-12 $$

Step 3:

The next rational root is $ x = 3 $

$$ \frac{ x^4-2x^3-11x^2+12x+36}{ x-3} = x^3+x^2-8x-12 $$

Step 4:

The next rational root is $ x = -2 $

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

Step 5:

The next rational root is $ x = 3 $

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

Step 6:

To find the last zero, solve equation $ x+2 = 0 $

$$ \begin{aligned} x+2 & = 0 \\[1 em] x & = -2 \end{aligned} $$

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