Find roots of polynomial $$ p(x) = s^3+5s^2-64s-140 $$
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= 7\\[1 em]s_2 &= -2\\[1 em]s_3 &= -10 \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ s = 7 } $ is a root of polynomial $ s^3+5s^2-64s-140 $.
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}{ 140 } $, with a single factor of 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70 and 140.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 140 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 14}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 28}{ 1} \pm \frac{ 35}{ 1} \pm \frac{ 70}{ 1} \pm \frac{ 140}{ 1} ~~ \end{aligned} $$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( 7 \right) = 0 $ so $ x = 7 $ 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}{ s-7 }$
$$ \frac{ s^3+5s^2-64s-140}{ s-7} = s^2+12s+20 $$Step 2:
The next rational root is $ s = 7 $
$$ \frac{ s^3+5s^2-64s-140}{ s-7} = s^2+12s+20 $$Step 3:
The next rational root is $ s = -2 $
$$ \frac{ s^2+12s+20}{ s+2} = s+10 $$Step 4:
To find the last zero, solve equation $ s+10 = 0 $
$$ \begin{aligned} s+10 & = 0 \\[1 em] s & = -10 \end{aligned} $$