Find roots of polynomial $$ p(x) = 8x^4+28x^3+9x^2-9x $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -3\\[1 em]x_3 &= 0.4114\\[1 em]x_4 &= -0.9114 \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ x }$ from $ 8x^4+28x^3+9x^2-9x $ and solve two separate equations:
$$ \begin{aligned} 8x^4+28x^3+9x^2-9x & = 0\\[1 em] \color{blue}{ x }\cdot ( 8x^3+28x^2+9x-9 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ 8x^3+28x^2+9x-9 & = 0 \end{aligned} $$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 $ 8x^3+28x^2+9x-9 $.
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}{ 9 } $, with factors of 1, 3 and 9.
The leading coefficient is $ \color{red}{ 8 }$, with factors of 1, 2, 4 and 8.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 9 }}{\text{ factors of 8 }} = \pm \dfrac{\text{ ( 1, 3, 9 ) }}{\text{ ( 1, 2, 4, 8 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 9}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 9}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 9}{ 8} ~~ \end{aligned} $$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 }$
$$ \frac{ 8x^3+28x^2+9x-9}{ x+3} = 8x^2+4x-3 $$Step 3:
The next rational root is $ x = -3 $
$$ \frac{ 8x^3+28x^2+9x-9}{ x+3} = 8x^2+4x-3 $$Step 4:
The solutions of $ 8x^2+4x-3 = 0 $ are: $ x = -\dfrac{ 1 }{ 4 }-\dfrac{\sqrt{ 7 }}{ 4 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 4 }+\dfrac{\sqrt{ 7 }}{ 4 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.