Find roots of polynomial $$ p(x) = -24a^3+28a^2+150a-175 $$
Answer
The roots of polynomial $ p(a) $ are:
$$ \begin{aligned}a_1 &= \frac{ 5 }{ 2 }\\[1 em]a_2 &= -\frac{ 5 }{ 2 }\\[1 em]a_3 &= \frac{ 7 }{ 6 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ a = \dfrac{ 5 }{ 2 } } $ is a root of polynomial $ -24a^3+28a^2+150a-175 $.
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}{ 175 } $, with factors of 1, 5, 7, 25, 35 and 175.
The leading coefficient is $ \color{red}{ 24 }$, with factors of 1, 2, 3, 4, 6, 8, 12 and 24.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 175 }}{\text{ factors of 24 }} = \pm \dfrac{\text{ ( 1, 5, 7, 25, 35, 175 ) }}{\text{ ( 1, 2, 3, 4, 6, 8, 12, 24 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 25}{ 1} \pm \frac{ 35}{ 1} \pm \frac{ 175}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 5}{ 2} \pm \frac{ 7}{ 2} \pm \frac{ 25}{ 2} \pm \frac{ 35}{ 2} \pm \frac{ 175}{ 2} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 5}{ 3} \pm \frac{ 7}{ 3} \pm \frac{ 25}{ 3} \pm \frac{ 35}{ 3} \pm \frac{ 175}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 5}{ 4} \pm \frac{ 7}{ 4} \pm \frac{ 25}{ 4} \pm \frac{ 35}{ 4} \pm \frac{ 175}{ 4} ~~ \pm \frac{ 1}{ 6} \pm \frac{ 5}{ 6} \pm \frac{ 7}{ 6} \pm \frac{ 25}{ 6} \pm \frac{ 35}{ 6} \pm \frac{ 175}{ 6} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 5}{ 8} \pm \frac{ 7}{ 8} \pm \frac{ 25}{ 8} \pm \frac{ 35}{ 8} \pm \frac{ 175}{ 8} ~~ \pm \frac{ 1}{ 12} \pm \frac{ 5}{ 12} \pm \frac{ 7}{ 12} \pm \frac{ 25}{ 12} \pm \frac{ 35}{ 12} \pm \frac{ 175}{ 12} ~~ \pm \frac{ 1}{ 24} \pm \frac{ 5}{ 24} \pm \frac{ 7}{ 24} \pm \frac{ 25}{ 24} \pm \frac{ 35}{ 24} \pm \frac{ 175}{ 24} ~~ \end{aligned} $$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{ 5 }{ 2 } \right) = 0 $ so $ x = \dfrac{ 5 }{ 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}{ 2a-5 }$
$$ \frac{ -24a^3+28a^2+150a-175}{ 2a-5} = -12a^2-16a+35 $$Step 2:
The next rational root is $ a = \dfrac{ 5 }{ 2 } $
$$ \frac{ -24a^3+28a^2+150a-175}{ 2a-5} = -12a^2-16a+35 $$Step 3:
The next rational root is $ a = -\dfrac{ 5 }{ 2 } $
$$ \frac{ -12a^2-16a+35}{ 2a+5} = -6a+7 $$Step 4:
To find the last zero, solve equation $ -6a+7 = 0 $
$$ \begin{aligned} -6a+7 & = 0 \\[1 em] -6 \cdot a & = -7 \\[1 em] a & = - \frac{ 7 }{ -6 } \\[1 em] a & = \frac{ 7 }{ 6 } \end{aligned} $$