Find roots of polynomial $$ p(x) = 16x^4-625 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 5 }{ 2 }\\[1 em]x_2 &= -\frac{ 5 }{ 2 }\\[1 em]x_3 &= \frac{ 5 }{ 2 }i\\[1 em]x_4 &= -\frac{ 5 }{ 2 }i \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 5 }{ 2 } } $ is a root of polynomial $ 16x^4-625 $.
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}{ 625 } $, with factors of 1, 5, 25, 125 and 625.
The leading coefficient is $ \color{red}{ 16 }$, with factors of 1, 2, 4, 8 and 16.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 625 }}{\text{ factors of 16 }} = \pm \dfrac{\text{ ( 1, 5, 25, 125, 625 ) }}{\text{ ( 1, 2, 4, 8, 16 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 125}{ 1} \pm \frac{ 625}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 125}{ 2} \pm \frac{ 625}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 25}{ 4} \pm \frac{ 125}{ 4} \pm \frac{ 625}{ 4}\\[ 1 em] \pm \frac{ 1}{ 8} & \pm \frac{ 5}{ 8} & \pm \frac{ 25}{ 8} & \pm \frac{ 125}{ 8} & \pm \frac{ 625}{ 8} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 5}{ 16} \pm \frac{ 25}{ 16} \pm \frac{ 125}{ 16} \pm \frac{ 625}{ 16} ~~ \end{aligned} $$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( \dfrac{ 5 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 5 }{ 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}{ 2x-5 }$
$$ \frac{ 16x^4-625}{ 2x-5} = 8x^3+20x^2+50x+125 $$Step 2:
The next rational root is $ x = \dfrac{ 5 }{ 2 } $
$$ \frac{ 16x^4-625}{ 2x-5} = 8x^3+20x^2+50x+125 $$Step 3:
The next rational root is $ x = -\dfrac{ 5 }{ 2 } $
$$ \frac{ 8x^3+20x^2+50x+125}{ 2x+5} = 4x^2+25 $$Step 4:
The solutions of $ 4x^2+25 = 0 $ are: $ x = \dfrac{ 5 }{ 2 } i ~ \text{and} ~ x = -\dfrac{ 5 }{ 2 } i $.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.