The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 4\\[1 em]x_2 &= -\frac{ 3 }{ 2 }\\[1 em]x_3 &= -\frac{ 5 }{ 2 } \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 4 } $ is a root of polynomial $ 4x^3-49x-60 $.
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}{ 60 } $, with factors of 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 60 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 10}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 20}{ 1} \pm \frac{ 30}{ 1} \pm \frac{ 60}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 4}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 10}{ 2} \pm \frac{ 12}{ 2} \pm \frac{ 15}{ 2} \pm \frac{ 20}{ 2} \pm \frac{ 30}{ 2} \pm \frac{ 60}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 4}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 10}{ 4} \pm \frac{ 12}{ 4} \pm \frac{ 15}{ 4} \pm \frac{ 20}{ 4} \pm \frac{ 30}{ 4} \pm \frac{ 60}{ 4} ~~ \end{aligned} $$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 }$
$$ \frac{ 4x^3-49x-60}{ x-4} = 4x^2+16x+15 $$Step 2:
The next rational root is $ x = 4 $
$$ \frac{ 4x^3-49x-60}{ x-4} = 4x^2+16x+15 $$Step 3:
The next rational root is $ x = -\dfrac{ 3 }{ 2 } $
$$ \frac{ 4x^2+16x+15}{ 2x+3} = 2x+5 $$Step 4:
To find the last zero, solve equation $ 2x+5 = 0 $
$$ \begin{aligned} 2x+5 & = 0 \\[1 em] 2 \cdot x & = -5 \\[1 em] x & = - \frac{ 5 }{ 2 } \end{aligned} $$