$$(x^2-49)(x^2+6x+9) = 0$$

$$ \begin{matrix}x_1 = -3 & x_2 = 7 & x_3 = -7 \end{matrix} $$

$ \color{blue}{ x^{4}+6x^{3}-40x^{2}-294x-441 } $ 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 (-441) are 1 3 7 9 21 49 63 147 441 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 49 }{ 1 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 147 }{ 1 } , ~ \pm \frac{ 441 }{ 1 } ~ $$

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(-3) = 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 + 3} $

$$ \frac{ x^{4}+6x^{3}-40x^{2}-294x-441 }{ \color{blue}{ x + 3 } } = x^{3}+3x^{2}-49x-147 $$

Polynomial $ x^{3}+3x^{2}-49x-147 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{3}+3x^{2}-49x-147 $

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

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