Find roots of polynomial $$ p(x) = 54+45x-24x^2-20x^3 $$
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \dfrac{ 3 }{ 2 }\\[1 em]x_2 &= -\dfrac{ 3 }{ 2 }\\[1 em]x_3 &= -\dfrac{ 6 }{ 5 } \end{aligned} $$Step 1:
Write polynomial in descending order
$$ \begin{aligned} 54+45x-24x^2-20x^3 & = 0\\[1 em] -20x^3-24x^2+45x+54 & = 0 \end{aligned} $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = \dfrac{ 3 }{ 2 } } $ is a root of polynomial $ -20x^3-24x^2+45x+54 $.
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}{ 54 } $, with factors of 1, 2, 3, 6, 9, 18, 27 and 54.
The leading coefficient is $ \color{red}{ 20 }$, with factors of 1, 2, 4, 5, 10 and 20.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 54 }}{\text{ factors of 20 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6, 9, 18, 27, 54 ) }}{\text{ ( 1, 2, 4, 5, 10, 20 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 18}{ 1} \pm \frac{ 27}{ 1} \pm \frac{ 54}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 18}{ 2} \pm \frac{ 27}{ 2} \pm \frac{ 54}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 18}{ 4} \pm \frac{ 27}{ 4} \pm \frac{ 54}{ 4} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 2}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 6}{ 5} \pm \frac{ 9}{ 5} \pm \frac{ 18}{ 5} \pm \frac{ 27}{ 5} \pm \frac{ 54}{ 5} ~~ \pm \frac{ 1}{ 10} \pm \frac{ 2}{ 10} \pm \frac{ 3}{ 10} \pm \frac{ 6}{ 10} \pm \frac{ 9}{ 10} \pm \frac{ 18}{ 10} \pm \frac{ 27}{ 10} \pm \frac{ 54}{ 10} ~~ \pm \frac{ 1}{ 20} \pm \frac{ 2}{ 20} \pm \frac{ 3}{ 20} \pm \frac{ 6}{ 20} \pm \frac{ 9}{ 20} \pm \frac{ 18}{ 20} \pm \frac{ 27}{ 20} \pm \frac{ 54}{ 20} ~~ \end{aligned} $$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( \dfrac{ 3 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 3 }{ 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}{ 2x-3 }$
$$ \frac{ -20x^3-24x^2+45x+54}{ 2x-3} = -10x^2-27x-18 $$Step 3:
The next rational root is $ x = \dfrac{ 3 }{ 2 } $
$$ \frac{ -20x^3-24x^2+45x+54}{ 2x-3} = -10x^2-27x-18 $$Step 4:
The next rational root is $ x = -\dfrac{ 3 }{ 2 } $
$$ \frac{ -10x^2-27x-18}{ 2x+3} = -5x-6 $$Step 5:
To find the last zero, solve equation $ -5x-6 = 0 $
$$ \begin{aligned} -5x-6 & = 0 \\[1 em] -5 \cdot x & = 6 \\[1 em] x & = \frac{ 6 }{ -5 } \end{aligned} $$