Find roots of polynomial $$ p(x) = 81x^4-81x^2+8 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 1 }{ 3 }\\[1 em]x_2 &= -\frac{ 1 }{ 3 }\\[1 em]x_3 &= \frac{ 2 \sqrt{ 2}}{ 3 }\\[1 em]x_4 &= -2 \frac{\sqrt{ 2 }}{ 3 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 1 }{ 3 } } $ is a root of polynomial $ 81x^4-81x^2+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}{ 81 }$, with factors of 1, 3, 9, 27 and 81.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 8 }}{\text{ factors of 81 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8 ) }}{\text{ ( 1, 3, 9, 27, 81 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 8}{ 3} ~~ \pm \frac{ 1}{ 9} \pm \frac{ 2}{ 9} \pm \frac{ 4}{ 9} \pm \frac{ 8}{ 9} ~~ \pm \frac{ 1}{ 27} \pm \frac{ 2}{ 27} \pm \frac{ 4}{ 27} \pm \frac{ 8}{ 27} ~~ \pm \frac{ 1}{ 81} \pm \frac{ 2}{ 81} \pm \frac{ 4}{ 81} \pm \frac{ 8}{ 81} ~~ \end{aligned} $$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 }{ 3 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 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}{ 3x-1 }$
$$ \frac{ 81x^4-81x^2+8}{ 3x-1} = 27x^3+9x^2-24x-8 $$Step 2:
The next rational root is $ x = \dfrac{ 1 }{ 3 } $
$$ \frac{ 81x^4-81x^2+8}{ 3x-1} = 27x^3+9x^2-24x-8 $$Step 3:
The next rational root is $ x = -\dfrac{ 1 }{ 3 } $
$$ \frac{ 27x^3+9x^2-24x-8}{ 3x+1} = 9x^2-8 $$Step 4:
The solutions of $ 9x^2-8 = 0 $ are: $ x = -2 \dfrac{\sqrt{ 2 }}{ 3 } ~ \text{and} ~ x = \dfrac{ 2 \sqrt{ 2}}{ 3 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.