$$(2x^2+13x+15)^3 = 0$$

$$ \begin{matrix}x_1 = -\dfrac{ 3 }{ 2 } & x_2 = -5 \\[1 em] \end{matrix} $$

$ \color{blue}{ 8x^{6}+156x^{5}+1194x^{4}+4537x^{3}+8955x^{2}+8775x+3375 } $ is a polynomial of degree 6. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factors of the leading coefficient ( 8 ) are 1 2 4 8 .The factors of the constant term (3375) are 1 3 5 9 15 25 27 45 75 125 135 225 375 675 1125 3375 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 8 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 5 }{ 8 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 4 } , ~ \pm \frac{ 9 }{ 8 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 4 } , ~ \pm \frac{ 15 }{ 8 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 25 }{ 2 } , ~ \pm \frac{ 25 }{ 4 } , ~ \pm \frac{ 25 }{ 8 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 27 }{ 4 } , ~ \pm \frac{ 27 }{ 8 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 45 }{ 2 } , ~ \pm \frac{ 45 }{ 4 } , ~ \pm \frac{ 45 }{ 8 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 75 }{ 2 } , ~ \pm \frac{ 75 }{ 4 } , ~ \pm \frac{ 75 }{ 8 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 125 }{ 2 } , ~ \pm \frac{ 125 }{ 4 } , ~ \pm \frac{ 125 }{ 8 } , ~ \pm \frac{ 135 }{ 1 } , ~ \pm \frac{ 135 }{ 2 } , ~ \pm \frac{ 135 }{ 4 } , ~ \pm \frac{ 135 }{ 8 } , ~ \pm \frac{ 225 }{ 1 } , ~ \pm \frac{ 225 }{ 2 } , ~ \pm \frac{ 225 }{ 4 } , ~ \pm \frac{ 225 }{ 8 } , ~ \pm \frac{ 375 }{ 1 } , ~ \pm \frac{ 375 }{ 2 } , ~ \pm \frac{ 375 }{ 4 } , ~ \pm \frac{ 375 }{ 8 } , ~ \pm \frac{ 675 }{ 1 } , ~ \pm \frac{ 675 }{ 2 } , ~ \pm \frac{ 675 }{ 4 } , ~ \pm \frac{ 675 }{ 8 } , ~ \pm \frac{ 1125 }{ 1 } , ~ \pm \frac{ 1125 }{ 2 } , ~ \pm \frac{ 1125 }{ 4 } , ~ \pm \frac{ 1125 }{ 8 } , ~ \pm \frac{ 3375 }{ 1 } , ~ \pm \frac{ 3375 }{ 2 } , ~ \pm \frac{ 3375 }{ 4 } , ~ \pm \frac{ 3375 }{ 8 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(-\frac{ 3 }{ 2 }) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{ 2 x + 3 } $

$$ \frac{ 8x^{6}+156x^{5}+1194x^{4}+4537x^{3}+8955x^{2}+8775x+3375 }{ \color{blue}{ 2x + 3 } } = 4x^{5}+72x^{4}+489x^{3}+1535x^{2}+2175x+1125 $$

Polynomial $ 4x^{5}+72x^{4}+489x^{3}+1535x^{2}+2175x+1125 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 4x^{5}+72x^{4}+489x^{3}+1535x^{2}+2175x+1125 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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