Find roots of polynomial $$ p(x) = x^5-8x^4+6x^3+108x^2-351x+324 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -4\\[1 em]x_3 &= 3\\[1 em]x_4 &= 3\\[1 em]x_5 &= 3 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ x^5-8x^4+6x^3+108x^2-351x+324 $.
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}{ 324 } $, with a single factor of 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162 and 324.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 324 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 324 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 54}{ 1} \pm \frac{ 81}{ 1} \pm \frac{ 108}{ 1} \pm \frac{ 162}{ 1} \pm \frac{ 324}{ 1} ~~ \end{aligned} $$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( 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 }$
$$ \frac{ x^5-8x^4+6x^3+108x^2-351x+324}{ x-3} = x^4-5x^3-9x^2+81x-108 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ x^5-8x^4+6x^3+108x^2-351x+324}{ x-3} = x^4-5x^3-9x^2+81x-108 $$Step 3:
The next rational root is $ x = -4 $
$$ \frac{ x^4-5x^3-9x^2+81x-108}{ x+4} = x^3-9x^2+27x-27 $$Step 4:
The next rational root is $ x = 3 $
$$ \frac{ x^3-9x^2+27x-27}{ x-3} = x^2-6x+9 $$Step 5:
The next rational root is $ x = 3 $
$$ \frac{ x^2-6x+9}{ x-3} = x-3 $$Step 6:
To find the last zero, solve equation $ x-3 = 0 $
$$ \begin{aligned} x-3 & = 0 \\[1 em] x & = 3 \end{aligned} $$