Find roots of polynomial $$ p(x) = 3x^5-4x^4+21x^3-28x^2-432x+576 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -3\\[1 em]x_3 &= \frac{ 4 }{ 3 }\\[1 em]x_4 &= 4i\\[1 em]x_5 &= -4i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ 3x^5-4x^4+21x^3-28x^2-432x+576 $.
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}{ 576 } $, with factors of 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 64, 72, 96, 144, 192, 288 and 576.
The leading coefficient is $ \color{red}{ 3 }$, with factors of 1 and 3.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 576 }}{\text{ factors of 3 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 64, 72, 96, 144, 192, 288, 576 ) }}{\text{ ( 1, 3 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 32}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 48}{ 1} \pm \frac{ 64}{ 1} \pm \frac{ 72}{ 1} \pm \frac{ 96}{ 1} \pm \frac{ 144}{ 1} \pm \frac{ 192}{ 1} \pm \frac{ 288}{ 1} \pm \frac{ 576}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 2}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 4}{ 3} \pm \frac{ 6}{ 3} \pm \frac{ 8}{ 3} \pm \frac{ 9}{ 3} \pm \frac{ 12}{ 3} \pm \frac{ 16}{ 3} \pm \frac{ 18}{ 3} \pm \frac{ 24}{ 3} \pm \frac{ 32}{ 3} \pm \frac{ 36}{ 3} \pm \frac{ 48}{ 3} \pm \frac{ 64}{ 3} \pm \frac{ 72}{ 3} \pm \frac{ 96}{ 3} \pm \frac{ 144}{ 3} \pm \frac{ 192}{ 3} \pm \frac{ 288}{ 3} \pm \frac{ 576}{ 3} ~~ \end{aligned} $$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 }$
$$ \frac{ 3x^5-4x^4+21x^3-28x^2-432x+576}{ x-3} = 3x^4+5x^3+36x^2+80x-192 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ 3x^5-4x^4+21x^3-28x^2-432x+576}{ x-3} = 3x^4+5x^3+36x^2+80x-192 $$Step 3:
The next rational root is $ x = -3 $
$$ \frac{ 3x^4+5x^3+36x^2+80x-192}{ x+3} = 3x^3-4x^2+48x-64 $$Step 4:
The next rational root is $ x = \dfrac{ 4 }{ 3 } $
$$ \frac{ 3x^3-4x^2+48x-64}{ 3x-4} = x^2+16 $$Step 5:
The solutions of $ x^2+16 = 0 $ are: $ x = 4 i ~ \text{and} ~ x = -4 i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.