Find roots of polynomial $$ p(x) = 2x^5+3x^4-30x^3-30x^2+88x-33 $$

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

$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -3\\[1 em]x_3 &= \frac{ 1 }{ 2 }\\[1 em]x_4 &= \sqrt{ 11 }\\[1 em]x_5 &= -\sqrt{ 11 } \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ 2x^5+3x^4-30x^3-30x^2+88x-33 $.

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}{ 33 } $, with factors of 1, 3, 11 and 33.

The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 33 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 3, 11, 33 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 11}{ 1} \pm \frac{ 33}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 11}{ 2} \pm \frac{ 33}{ 2} ~~ \end{aligned} $$

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 }$

$$ \frac{ 2x^5+3x^4-30x^3-30x^2+88x-33}{ x-1} = 2x^4+5x^3-25x^2-55x+33 $$

Step 2:

The next rational root is $ x = 1 $

$$ \frac{ 2x^5+3x^4-30x^3-30x^2+88x-33}{ x-1} = 2x^4+5x^3-25x^2-55x+33 $$

Step 3:

The next rational root is $ x = -3 $

$$ \frac{ 2x^4+5x^3-25x^2-55x+33}{ x+3} = 2x^3-x^2-22x+11 $$

Step 4:

The next rational root is $ x = \dfrac{ 1 }{ 2 } $

$$ \frac{ 2x^3-x^2-22x+11}{ 2x-1} = x^2-11 $$

Step 5:

The solutions of $ x^2-11 = 0 $ are: $ x = -\sqrt{ 11 } ~ \text{and} ~ x = \sqrt{ 11 }$.

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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