Find roots of polynomial $$ p(x) = 15x^3-107x^2-247x-45 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 9\\[1 em]x_2 &= -\dfrac{ 5 }{ 3 }\\[1 em]x_3 &= -\dfrac{ 1 }{ 5 } \end{aligned} $$Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 9 } $ is a root of polynomial $ 15x^3-107x^2-247x-45 $.
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}{ 45 } $, with factors of 1, 3, 5, 9, 15 and 45.
The leading coefficient is $ \color{red}{ 15 }$, with factors of 1, 3, 5 and 15.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 45 }}{\text{ factors of 15 }} = \pm \dfrac{\text{ ( 1, 3, 5, 9, 15, 45 ) }}{\text{ ( 1, 3, 5, 15 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 15}{ 1} \pm \frac{ 45}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 3}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 9}{ 3} \pm \frac{ 15}{ 3} \pm \frac{ 45}{ 3} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 5}{ 5} \pm \frac{ 9}{ 5} \pm \frac{ 15}{ 5} \pm \frac{ 45}{ 5}\\[ 1 em] \pm \frac{ 1}{ 15} & \pm \frac{ 3}{ 15} & \pm \frac{ 5}{ 15} & \pm \frac{ 9}{ 15} & \pm \frac{ 15}{ 15} & \pm \frac{ 45}{ 15} ~~ \end{aligned} $$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( 9 \right) = 0 $ so $ x = 9 $ 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-9 }$
$$ \frac{ 15x^3-107x^2-247x-45}{ x-9} = 15x^2+28x+5 $$Step 2:
The next rational root is $ x = 9 $
$$ \frac{ 15x^3-107x^2-247x-45}{ x-9} = 15x^2+28x+5 $$Step 3:
The next rational root is $ x = -\dfrac{ 5 }{ 3 } $
$$ \frac{ 15x^2+28x+5}{ 3x+5} = 5x+1 $$Step 4:
To find the last zero, solve equation $ 5x+1 = 0 $
$$ \begin{aligned} 5x+1 & = 0 \\[1 em] 5 \cdot x & = -1 \\[1 em] x & = - \frac{ 1 }{ 5 } \end{aligned} $$