Find roots of polynomial $$ p(x) = -48x^5+128x^4-24x^3+64x^2 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= \frac{ 8 }{ 3 }\\[1 em]x_3 &= \frac{\sqrt{ 2 }}{ 2 }i\\[1 em]x_4 &= - \frac{\sqrt{ 2 }}{ 2 }i \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ -8x^2 }$ from $ -48x^5+128x^4-24x^3+64x^2 $ and solve two separate equations:
$$ \begin{aligned} -48x^5+128x^4-24x^3+64x^2 & = 0\\[1 em] \color{blue}{ -8x^2 }\cdot ( 6x^3-16x^2+3x-8 ) & = 0 \\[1 em] \color{blue}{ -8x^2 = 0} ~~ \text{or} ~~ 6x^3-16x^2+3x-8 & = 0 \end{aligned} $$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{ 8 }{ 3 } } $ is a root of polynomial $ 6x^3-16x^2+3x-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}{ 6 }$, with factors of 1, 2, 3 and 6.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 8 }}{\text{ factors of 6 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8 ) }}{\text{ ( 1, 2, 3, 6 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 8}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 8}{ 3} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 2}{ 6} \pm \frac{ 4}{ 6} \pm \frac{ 8}{ 6}\\[ 1 em] \end{aligned} $$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{ 8 }{ 3 } \right) = 0 $ so $ x = \dfrac{ 8 }{ 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-8 }$
$$ \frac{ 6x^3-16x^2+3x-8}{ 3x-8} = 2x^2+1 $$Step 3:
The next rational root is $ x = \dfrac{ 8 }{ 3 } $
$$ \frac{ 6x^3-16x^2+3x-8}{ 3x-8} = 2x^2+1 $$Step 4:
The solutions of $ 2x^2+1 = 0 $ are: $ x = \dfrac{\sqrt{ 2 }}{ 2 } i ~ \text{and} ~ x = - \dfrac{\sqrt{ 2 }}{ 2 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.