Find roots of polynomial $$ p(x) = -4x^5+64x^4-212x^3-424x^2-784x-2720 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -2\\[1 em]x_2 &= 2i\\[1 em]x_3 &= -2i\\[1 em]x_4 &= 9+2i\\[1 em]x_5 &= 9-2i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -2 } $ is a root of polynomial $ -4x^5+64x^4-212x^3-424x^2-784x-2720 $.
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}{ 2720 } $, with factors of 1, 2, 4, 5, 8, 10, 16, 17, 20, 32, 34, 40, 68, 80, 85, 136, 160, 170, 272, 340, 544, 680, 1360 and 2720.
The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 2720 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 8, 10, 16, 17, 20, 32, 34, 40, 68, 80, 85, 136, 160, 170, 272, 340, 544, 680, 1360, 2720 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 17}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 34}{ 1} \pm \frac{ 40}{ 1} \pm \frac{ 68}{ 1} \pm \frac{ 80}{ 1} \pm \frac{ 85}{ 1} \pm \frac{ 136}{ 1} \pm \frac{ 160}{ 1} \pm \frac{ 170}{ 1} \pm \frac{ 272}{ 1} \pm \frac{ 340}{ 1} \pm \frac{ 544}{ 1} \pm \frac{ 680}{ 1} \pm \frac{ 1360}{ 1} \pm \frac{ 2720}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 16}{ 2} \pm \frac{ 17}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 32}{ 2} \pm \frac{ 34}{ 2} \pm \frac{ 40}{ 2} \pm \frac{ 68}{ 2} \pm \frac{ 80}{ 2} \pm \frac{ 85}{ 2} \pm \frac{ 136}{ 2} \pm \frac{ 160}{ 2} \pm \frac{ 170}{ 2} \pm \frac{ 272}{ 2} \pm \frac{ 340}{ 2} \pm \frac{ 544}{ 2} \pm \frac{ 680}{ 2} \pm \frac{ 1360}{ 2} \pm \frac{ 2720}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 16}{ 4} \pm \frac{ 17}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 32}{ 4} \pm \frac{ 34}{ 4} \pm \frac{ 40}{ 4} \pm \frac{ 68}{ 4} \pm \frac{ 80}{ 4} \pm \frac{ 85}{ 4} \pm \frac{ 136}{ 4} \pm \frac{ 160}{ 4} \pm \frac{ 170}{ 4} \pm \frac{ 272}{ 4} \pm \frac{ 340}{ 4} \pm \frac{ 544}{ 4} \pm \frac{ 680}{ 4} \pm \frac{ 1360}{ 4} \pm \frac{ 2720}{ 4} ~~ \end{aligned} $$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( -2 \right) = 0 $ so $ x = -2 $ 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+2 }$
$$ \frac{ -4x^5+64x^4-212x^3-424x^2-784x-2720}{ x+2} = -4x^4+72x^3-356x^2+288x-1360 $$Step 2:
The next rational root is $ x = -2 $
$$ \frac{ -4x^5+64x^4-212x^3-424x^2-784x-2720}{ x+2} = -4x^4+72x^3-356x^2+288x-1360 $$Step 3:
Polynomial $ -4x^4+72x^3-356x^2+288x-1360 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.