$$(4x+2)^3 = 0$$
Answer
Explanation
$ \color{blue}{ 64x^{3}+96x^{2}+48x+8 } $ is a polynomial of degree 3. 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 ( 64 ) are 1 2 4 8 16 32 64 .The factors of the constant term (8) are 1 2 4 8 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 16 } , ~ \pm \frac{ 1 }{ 32 } , ~ \pm \frac{ 1 }{ 64 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 8 } , ~ \pm \frac{ 2 }{ 16 } , ~ \pm \frac{ 2 }{ 32 } , ~ \pm \frac{ 2 }{ 64 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 4 }{ 8 } , ~ \pm \frac{ 4 }{ 16 } , ~ \pm \frac{ 4 }{ 32 } , ~ \pm \frac{ 4 }{ 64 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 4 } , ~ \pm \frac{ 8 }{ 8 } , ~ \pm \frac{ 8 }{ 16 } , ~ \pm \frac{ 8 }{ 32 } , ~ \pm \frac{ 8 }{ 64 } ~ $$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{ 1 }{ 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 + 1 } $
$$ \frac{ 64x^{3}+96x^{2}+48x+8 }{ \color{blue}{ 2x + 1 } } = 32x^{2}+32x+8 $$Polynomial $ 32x^{2}+32x+8 $ can be used to find the remaining roots.
$ \color{blue}{ 32x^{2}+32x+8 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.