Find roots of polynomial $$ p(x) = x^3+24x^2+192x+512 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -8\\[1 em]x_2 &= -8\\[1 em]x_3 &= -8 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -8 } $ is a root of polynomial $ x^3+24x^2+192x+512 $.
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}{ 512 } $, with a single factor of 1, 2, 4, 8, 16, 32, 64, 128, 256 and 512.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 512 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 128}{ 1} \pm \frac{ 256}{ 1} \pm \frac{ 512}{ 1} ~~ \end{aligned} $$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( -8 \right) = 0 $ so $ x = -8 $ 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+8 }$
$$ \frac{ x^3+24x^2+192x+512}{ x+8} = x^2+16x+64 $$Step 2:
The next rational root is $ x = -8 $
$$ \frac{ x^3+24x^2+192x+512}{ x+8} = x^2+16x+64 $$Step 3:
The next rational root is $ x = -8 $
$$ \frac{ x^2+16x+64}{ x+8} = x+8 $$Step 4:
To find the last zero, solve equation $ x+8 = 0 $
$$ \begin{aligned} x+8 & = 0 \\[1 em] x & = -8 \end{aligned} $$