Find roots of polynomial $$ p(x) = 15a^3-5a^2+15a-5 $$

The roots of polynomial $ p(a) $ are:

$$ \begin{aligned}a_1 &= \frac{ 1 }{ 3 }\\[1 em]a_2 &= i\\[1 em]a_3 &= -i \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ a = \dfrac{ 1 }{ 3 } } $ is a root of polynomial $ 15a^3-5a^2+15a-5 $.

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}{ 5 } $, with factors of 1 and 5.

The leading coefficient is $ \color{red}{ 15 }$, with factors of 1, 3, 5 and 15.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 5 }}{\text{ factors of 15 }} = \pm \dfrac{\text{ ( 1, 5 ) }}{\text{ ( 1, 3, 5, 15 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 5}{ 1} ~~ \pm \frac{ 1}{ 3} \pm \frac{ 5}{ 3} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 5}{ 5} ~~ \pm \frac{ 1}{ 15} \pm \frac{ 5}{ 15}\\[ 1 em] \end{aligned} $$

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{ 1 }{ 3 } \right) = 0 $ so $ x = \dfrac{ 1 }{ 3 } $ 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}{ 3a-1 }$

$$ \frac{ 15a^3-5a^2+15a-5}{ 3a-1} = 5a^2+5 $$

Step 2:

The next rational root is $ a = \dfrac{ 1 }{ 3 } $

$$ \frac{ 15a^3-5a^2+15a-5}{ 3a-1} = 5a^2+5 $$

Step 3:

The solutions of $ 5a^2+5 = 0 $ are: $ a = i ~ \text{and} ~ a = -i$.

You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.

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