Find roots of polynomial $$ p(x) = 16x^3+10x^2-486x+180 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 5\\[1 em]x_2 &= -6\\[1 em]x_3 &= \dfrac{ 3 }{ 8 } \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 5 } $ is a root of polynomial $ 16x^3+10x^2-486x+180 $.
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}{ 180 } $, with factors of 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90 and 180.
The leading coefficient is $ \color{red}{ 16 }$, with factors of 1, 2, 4, 8 and 16.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 180 }}{\text{ factors of 16 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 ) }}{\text{ ( 1, 2, 4, 8, 16 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 36}{ 1} \pm \frac{ 45}{ 1} \pm \frac{ 60}{ 1} \pm \frac{ 90}{ 1} \pm \frac{ 180}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 18}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 30}{ 2} \pm \frac{ 36}{ 2} \pm \frac{ 45}{ 2} \pm \frac{ 60}{ 2} \pm \frac{ 90}{ 2} \pm \frac{ 180}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 12}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 18}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 30}{ 4} \pm \frac{ 36}{ 4} \pm \frac{ 45}{ 4} \pm \frac{ 60}{ 4} \pm \frac{ 90}{ 4} \pm \frac{ 180}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 2}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 4}{ 8} \pm \frac{ 5}{ 8} \pm \frac{ 6}{ 8} \pm \frac{ 9}{ 8} \pm \frac{ 10}{ 8} \pm \frac{ 12}{ 8} \pm \frac{ 15}{ 8} \pm \frac{ 18}{ 8} \pm \frac{ 20}{ 8} \pm \frac{ 30}{ 8} \pm \frac{ 36}{ 8} \pm \frac{ 45}{ 8} \pm \frac{ 60}{ 8} \pm \frac{ 90}{ 8} \pm \frac{ 180}{ 8} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 2}{ 16} \pm \frac{ 3}{ 16} \pm \frac{ 4}{ 16} \pm \frac{ 5}{ 16} \pm \frac{ 6}{ 16} \pm \frac{ 9}{ 16} \pm \frac{ 10}{ 16} \pm \frac{ 12}{ 16} \pm \frac{ 15}{ 16} \pm \frac{ 18}{ 16} \pm \frac{ 20}{ 16} \pm \frac{ 30}{ 16} \pm \frac{ 36}{ 16} \pm \frac{ 45}{ 16} \pm \frac{ 60}{ 16} \pm \frac{ 90}{ 16} \pm \frac{ 180}{ 16} ~~ \end{aligned} $$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( 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 }$
$$ \frac{ 16x^3+10x^2-486x+180}{ x-5} = 16x^2+90x-36 $$Step 2:
The next rational root is $ x = 5 $
$$ \frac{ 16x^3+10x^2-486x+180}{ x-5} = 16x^2+90x-36 $$Step 3:
The next rational root is $ x = -6 $
$$ \frac{ 16x^2+90x-36}{ x+6} = 16x-6 $$Step 4:
To find the last zero, solve equation $ 16x-6 = 0 $
$$ \begin{aligned} 16x-6 & = 0 \\[1 em] 16 \cdot x & = 6 \\[1 em] x & = \frac{ 6 }{ 16 } \\[1 em] x & = \frac{ 6 : 2 }{ 16 : 2} \\[1 em] x & = \frac{ 3 }{ 8 } \end{aligned} $$