Find roots of polynomial $$ p(x) = x^8-18x^6-320x^4+11x^7-472x^5+6576x^3-27648x+7776x^2-41472 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= 4\\[1 em]x_3 &= -2\\[1 em]x_4 &= -6\\[1 em]x_5 &= 4\\[1 em]x_6 &= -2\\[1 em]x_7 &= -6\\[1 em]x_8 &= -6 \end{aligned} $$Explanation
Step 1:
Write polynomial in descending order
$$ \begin{aligned} x^8-18x^6-320x^4+11x^7-472x^5+6576x^3-27648x+7776x^2-41472 & = 0\\[1 em] x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472 $.
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}{ 41472 } $, with a single factor of 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 256, 288, 324, 384, 432, 512, 576, 648, 768, 864, 1152, 1296, 1536, 1728, 2304, 2592, 3456, 4608, 5184, 6912, 10368, 13824, 20736 and 41472.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 41472 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 256, 288, 324, 384, 432, 512, 576, 648, 768, 864, 1152, 1296, 1536, 1728, 2304, 2592, 3456, 4608, 5184, 6912, 10368, 13824, 20736, 41472 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 48}{ 1} \pm \frac{ 54}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 72}{ 1} \pm \frac{ 81}{ 1} \pm \frac{ 96}{ 1} \pm \frac{ 108}{ 1} \pm \frac{ 128}{ 1} \pm \frac{ 144}{ 1} \pm \frac{ 162}{ 1} \pm \frac{ 192}{ 1} \pm \frac{ 216}{ 1} \pm \frac{ 256}{ 1} \pm \frac{ 288}{ 1} \pm \frac{ 324}{ 1} \pm \frac{ 384}{ 1} \pm \frac{ 432}{ 1} \pm \frac{ 512}{ 1} \pm \frac{ 576}{ 1} \pm \frac{ 648}{ 1} \pm \frac{ 768}{ 1} \pm \frac{ 864}{ 1} \pm \frac{ 1152}{ 1} \pm \frac{ 1296}{ 1} \pm \frac{ 1536}{ 1} \pm \frac{ 1728}{ 1} \pm \frac{ 2304}{ 1} \pm \frac{ 2592}{ 1} \pm \frac{ 3456}{ 1} \pm \frac{ 4608}{ 1} \pm \frac{ 5184}{ 1} \pm \frac{ 6912}{ 1} \pm \frac{ 10368}{ 1} \pm \frac{ 13824}{ 1} \pm \frac{ 20736}{ 1} \pm \frac{ 41472}{ 1} ~~ \end{aligned} $$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 }$
$$ \frac{ x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472}{ x-3} = x^7+14x^6+24x^5-400x^4-1520x^3+2016x^2+13824x+13824 $$Step 3:
The next rational root is $ x = 3 $
$$ \frac{ x^8+11x^7-18x^6-472x^5-320x^4+6576x^3+7776x^2-27648x-41472}{ x-3} = x^7+14x^6+24x^5-400x^4-1520x^3+2016x^2+13824x+13824 $$Step 4:
The next rational root is $ x = 4 $
$$ \frac{ x^7+14x^6+24x^5-400x^4-1520x^3+2016x^2+13824x+13824}{ x-4} = x^6+18x^5+96x^4-16x^3-1584x^2-4320x-3456 $$Step 5:
The next rational root is $ x = -2 $
$$ \frac{ x^6+18x^5+96x^4-16x^3-1584x^2-4320x-3456}{ x+2} = x^5+16x^4+64x^3-144x^2-1296x-1728 $$Step 6:
The next rational root is $ x = -6 $
$$ \frac{ x^5+16x^4+64x^3-144x^2-1296x-1728}{ x+6} = x^4+10x^3+4x^2-168x-288 $$Step 7:
The next rational root is $ x = 4 $
$$ \frac{ x^4+10x^3+4x^2-168x-288}{ x-4} = x^3+14x^2+60x+72 $$Step 8:
The next rational root is $ x = -2 $
$$ \frac{ x^3+14x^2+60x+72}{ x+2} = x^2+12x+36 $$Step 9:
The next rational root is $ x = -6 $
$$ \frac{ x^2+12x+36}{ x+6} = x+6 $$Step 10:
To find the last zero, solve equation $ x+6 = 0 $
$$ \begin{aligned} x+6 & = 0 \\[1 em] x & = -6 \end{aligned} $$