Find roots of polynomial $$ p(x) = 8x^4-19x^3-95x^2+142x+120 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= 4\\[1 em]x_3 &= -3\\[1 em]x_4 &= -\frac{ 5 }{ 8 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 8x^4-19x^3-95x^2+142x+120 $.
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}{ 120 } $, with factors of 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
The leading coefficient is $ \color{red}{ 8 }$, with factors of 1, 2, 4 and 8.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 120 }}{\text{ factors of 8 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ) }}{\text{ ( 1, 2, 4, 8 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 40}{ 1} \pm \frac{ 60}{ 1} \pm \frac{ 120}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 8}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 24}{ 2} \pm \frac{ 30}{ 2} \pm \frac{ 40}{ 2} \pm \frac{ 60}{ 2} \pm \frac{ 120}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 8}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 12}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 24}{ 4} \pm \frac{ 30}{ 4} \pm \frac{ 40}{ 4} \pm \frac{ 60}{ 4} \pm \frac{ 120}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 2}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 4}{ 8} \pm \frac{ 5}{ 8} \pm \frac{ 6}{ 8} \pm \frac{ 8}{ 8} \pm \frac{ 10}{ 8} \pm \frac{ 12}{ 8} \pm \frac{ 15}{ 8} \pm \frac{ 20}{ 8} \pm \frac{ 24}{ 8} \pm \frac{ 30}{ 8} \pm \frac{ 40}{ 8} \pm \frac{ 60}{ 8} \pm \frac{ 120}{ 8}\\[ 1 em] \end{aligned} $$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( 2 \right) = 0 $ so $ x = 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}{ x-2 }$
$$ \frac{ 8x^4-19x^3-95x^2+142x+120}{ x-2} = 8x^3-3x^2-101x-60 $$Step 2:
The next rational root is $ x = 2 $
$$ \frac{ 8x^4-19x^3-95x^2+142x+120}{ x-2} = 8x^3-3x^2-101x-60 $$Step 3:
The next rational root is $ x = 4 $
$$ \frac{ 8x^3-3x^2-101x-60}{ x-4} = 8x^2+29x+15 $$Step 4:
The next rational root is $ x = -3 $
$$ \frac{ 8x^2+29x+15}{ x+3} = 8x+5 $$Step 5:
To find the last zero, solve equation $ 8x+5 = 0 $
$$ \begin{aligned} 8x+5 & = 0 \\[1 em] 8 \cdot x & = -5 \\[1 em] x & = - \frac{ 5 }{ 8 } \end{aligned} $$