Find roots of polynomial $$ p(x) = x^3+25x^2+199x+495 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -5\\[1 em]x_2 &= -9\\[1 em]x_3 &= -11 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -5 } $ is a root of polynomial $ x^3+25x^2+199x+495 $.
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}{ 495 } $, with a single factor of 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165 and 495.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 495 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 11}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 33}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 55}{ 1} \pm \frac{ 99}{ 1} \pm \frac{ 165}{ 1} \pm \frac{ 495}{ 1} ~~ \end{aligned} $$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( -5 \right) = 0 $ so $ x = -5 $ 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+5 }$
$$ \frac{ x^3+25x^2+199x+495}{ x+5} = x^2+20x+99 $$Step 2:
The next rational root is $ x = -5 $
$$ \frac{ x^3+25x^2+199x+495}{ x+5} = x^2+20x+99 $$Step 3:
The next rational root is $ x = -9 $
$$ \frac{ x^2+20x+99}{ x+9} = x+11 $$Step 4:
To find the last zero, solve equation $ x+11 = 0 $
$$ \begin{aligned} x+11 & = 0 \\[1 em] x & = -11 \end{aligned} $$