Find roots of polynomial $$ p(x) = x^3-13x^2-9x+405 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 9\\[1 em]x_2 &= -5\\[1 em]x_3 &= 9 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 9 } $ is a root of polynomial $ x^3-13x^2-9x+405 $.
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}{ 405 } $, with a single factor of 1, 3, 5, 9, 15, 27, 45, 81, 135 and 405.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 405 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 15, 27, 45, 81, 135, 405 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 81}{ 1} \pm \frac{ 135}{ 1} \pm \frac{ 405}{ 1} ~~ \end{aligned} $$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( 9 \right) = 0 $ so $ x = 9 $ 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-9 }$
$$ \frac{ x^3-13x^2-9x+405}{ x-9} = x^2-4x-45 $$Step 2:
The next rational root is $ x = 9 $
$$ \frac{ x^3-13x^2-9x+405}{ x-9} = x^2-4x-45 $$Step 3:
The next rational root is $ x = -5 $
$$ \frac{ x^2-4x-45}{ x+5} = x-9 $$Step 4:
To find the last zero, solve equation $ x-9 = 0 $
$$ \begin{aligned} x-9 & = 0 \\[1 em] x & = 9 \end{aligned} $$