Find roots of polynomial $$ p(x) = 2x^4+23x^3+60x^2-125x-500 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -4\\[1 em]x_2 &= -5\\[1 em]x_3 &= \frac{ 5 }{ 2 }\\[1 em]x_4 &= -5 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -4 } $ is a root of polynomial $ 2x^4+23x^3+60x^2-125x-500 $.
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}{ 500 } $, with factors of 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250 and 500.
The leading coefficient is $ \color{red}{ 2 }$, with factors of 1 and 2.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 500 }}{\text{ factors of 2 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500 ) }}{\text{ ( 1, 2 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 50}{ 1} \pm \frac{ 100}{ 1} \pm \frac{ 125}{ 1} \pm \frac{ 250}{ 1} \pm \frac{ 500}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 50}{ 2} \pm \frac{ 100}{ 2} \pm \frac{ 125}{ 2} \pm \frac{ 250}{ 2} \pm \frac{ 500}{ 2} ~~ \end{aligned} $$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( -4 \right) = 0 $ so $ x = -4 $ 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+4 }$
$$ \frac{ 2x^4+23x^3+60x^2-125x-500}{ x+4} = 2x^3+15x^2-125 $$Step 2:
The next rational root is $ x = -4 $
$$ \frac{ 2x^4+23x^3+60x^2-125x-500}{ x+4} = 2x^3+15x^2-125 $$Step 3:
The next rational root is $ x = -5 $
$$ \frac{ 2x^3+15x^2-125}{ x+5} = 2x^2+5x-25 $$Step 4:
The next rational root is $ x = \dfrac{ 5 }{ 2 } $
$$ \frac{ 2x^2+5x-25}{ 2x-5} = x+5 $$Step 5:
To find the last zero, solve equation $ x+5 = 0 $
$$ \begin{aligned} x+5 & = 0 \\[1 em] x & = -5 \end{aligned} $$