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

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

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{ 1 }{ 4 }\\[1 em]x_3 &= \sqrt{ 2 }i\\[1 em]x_4 &= -\sqrt{ 2 }i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ 4x^4-x^3+8x^2-2x $ and solve two separate equations:

$$ \begin{aligned} 4x^4-x^3+8x^2-2x & = 0\\[1 em] \color{blue}{ x }\cdot ( 4x^3-x^2+8x-2 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 4x^3-x^2+8x-2 & = 0 \end{aligned} $$

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 = \dfrac{ 1 }{ 4 } } $ is a root of polynomial $ 4x^3-x^2+8x-2 $.

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

The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.

The POSSIBLE zeroes are:

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

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( \dfrac{ 1 }{ 4 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 4 } $ 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}{ 4x-1 }$

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

Step 3:

The next rational root is $ x = \dfrac{ 1 }{ 4 } $

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

Step 4:

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