$$x^2(x+1)^2+4(x+1)^3 = 0$$
Answer
Explanation
$ \color{blue}{ x^{4}+6x^{3}+13x^{2}+12x+4 } $ is a polynomial of degree 4. 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (4) are 1 2 4 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } ~ $$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(-1) = 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}{x + 1} $
$$ \frac{ x^{4}+6x^{3}+13x^{2}+12x+4 }{ \color{blue}{ x + 1 } } = x^{3}+5x^{2}+8x+4 $$Polynomial $ x^{3}+5x^{2}+8x+4 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}+5x^{2}+8x+4 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.