Find roots of polynomial $$ p(x) = 20x^4+10x^2-2552x+10 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 5\\[1 em]x_2 &= 0.0039\\[1 em]x_3 &= -2.502+4.3886i\\[1 em]x_4 &= -2.502-4.3886i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 5 } $ is a root of polynomial $ 20x^4+10x^2-2552x+10 $.
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}{ 10 } $, with factors of 1, 2, 5 and 10.
The leading coefficient is $ \color{red}{ 20 }$, with factors of 1, 2, 4, 5, 10 and 20.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 10 }}{\text{ factors of 20 }} = \pm \dfrac{\text{ ( 1, 2, 5, 10 ) }}{\text{ ( 1, 2, 4, 5, 10, 20 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 10}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 10}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 10}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 5}{ 5} \pm \frac{ 10}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 2}{ 10} \pm \frac{ 5}{ 10} \pm \frac{ 10}{ 10} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 2}{ 20} \pm \frac{ 5}{ 20} \pm \frac{ 10}{ 20} ~~ \end{aligned} $$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( 5 \right) = 0 $ so $ x = 5 $ 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-5 }$
$$ \frac{ 20x^4+10x^2-2552x+10}{ x-5} = 20x^3+100x^2+510x-2 $$Step 2:
The next rational root is $ x = 5 $
$$ \frac{ 20x^4+10x^2-2552x+10}{ x-5} = 20x^3+100x^2+510x-2 $$Step 3:
Polynomial $ 20x^3+100x^2+510x-2 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.