Find roots of polynomial $$ p(x) = 4x^6-435x^4+3491x^2+900 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= 10\\[1 em]x_3 &= -3\\[1 em]x_4 &= -10\\[1 em]x_5 &= \frac{ 1 }{ 2 }i\\[1 em]x_6 &= -\frac{ 1 }{ 2 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ 4x^6-435x^4+3491x^2+900 $.
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}{ 900 } $, with factors of 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450 and 900.
The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 900 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450, 900 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 60}{ 1} \pm \frac{ 75}{ 1} \pm \frac{ 90}{ 1} \pm \frac{ 100}{ 1} \pm \frac{ 150}{ 1} \pm \frac{ 180}{ 1} \pm \frac{ 225}{ 1} \pm \frac{ 300}{ 1} \pm \frac{ 450}{ 1} \pm \frac{ 900}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 18}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 30}{ 2} \pm \frac{ 36}{ 2} \pm \frac{ 45}{ 2} \pm \frac{ 50}{ 2} \pm \frac{ 60}{ 2} \pm \frac{ 75}{ 2} \pm \frac{ 90}{ 2} \pm \frac{ 100}{ 2} \pm \frac{ 150}{ 2} \pm \frac{ 180}{ 2} \pm \frac{ 225}{ 2} \pm \frac{ 300}{ 2} \pm \frac{ 450}{ 2} \pm \frac{ 900}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 12}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 18}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 25}{ 4} \pm \frac{ 30}{ 4} \pm \frac{ 36}{ 4} \pm \frac{ 45}{ 4} \pm \frac{ 50}{ 4} \pm \frac{ 60}{ 4} \pm \frac{ 75}{ 4} \pm \frac{ 90}{ 4} \pm \frac{ 100}{ 4} \pm \frac{ 150}{ 4} \pm \frac{ 180}{ 4} \pm \frac{ 225}{ 4} \pm \frac{ 300}{ 4} \pm \frac{ 450}{ 4} \pm \frac{ 900}{ 4} ~~ \end{aligned} $$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( 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 }$
$$ \frac{ 4x^6-435x^4+3491x^2+900}{ x-3} = 4x^5+12x^4-399x^3-1197x^2-100x-300 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ 4x^6-435x^4+3491x^2+900}{ x-3} = 4x^5+12x^4-399x^3-1197x^2-100x-300 $$Step 3:
The next rational root is $ x = 10 $
$$ \frac{ 4x^5+12x^4-399x^3-1197x^2-100x-300}{ x-10} = 4x^4+52x^3+121x^2+13x+30 $$Step 4:
The next rational root is $ x = -3 $
$$ \frac{ 4x^4+52x^3+121x^2+13x+30}{ x+3} = 4x^3+40x^2+x+10 $$Step 5:
The next rational root is $ x = -10 $
$$ \frac{ 4x^3+40x^2+x+10}{ x+10} = 4x^2+1 $$Step 6:
The solutions of $ 4x^2+1 = 0 $ are: $ x = \dfrac{ 1 }{ 2 } i ~ \text{and} ~ x = -\dfrac{ 1 }{ 2 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.