Find roots of polynomial $$ p(x) = x^4+8x^3+15x^2-8x-16 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= -1\\[1 em]x_3 &= -4\\[1 em]x_4 &= -4 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ x^4+8x^3+15x^2-8x-16 $.
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}{ 16 } $, with a single factor of 1, 2, 4, 8 and 16.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 16 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 4, 8, 16 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 16}{ 1} ~~ \end{aligned} $$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( 1 \right) = 0 $ so $ x = 1 $ 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-1 }$
$$ \frac{ x^4+8x^3+15x^2-8x-16}{ x-1} = x^3+9x^2+24x+16 $$Step 2:
The next rational root is $ x = 1 $
$$ \frac{ x^4+8x^3+15x^2-8x-16}{ x-1} = x^3+9x^2+24x+16 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ x^3+9x^2+24x+16}{ x+1} = x^2+8x+16 $$Step 4:
The next rational root is $ x = -4 $
$$ \frac{ x^2+8x+16}{ x+4} = x+4 $$Step 5:
To find the last zero, solve equation $ x+4 = 0 $
$$ \begin{aligned} x+4 & = 0 \\[1 em] x & = -4 \end{aligned} $$