$ \color{blue}{ x^{4}-2x^{3}-67x^{2}+68x+672 } $ 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 (672) are 1 2 3 4 6 7 8 12 14 16 21 24 28 32 42 48 56 84 96 112 168 224 336 672 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 28 }{ 1 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 56 }{ 1 } , ~ \pm \frac{ 84 }{ 1 } , ~ \pm \frac{ 96 }{ 1 } , ~ \pm \frac{ 112 }{ 1 } , ~ \pm \frac{ 168 }{ 1 } , ~ \pm \frac{ 224 }{ 1 } , ~ \pm \frac{ 336 }{ 1 } , ~ \pm \frac{ 672 }{ 1 } ~ $$

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} $

$$ \frac{ x^{4}-2x^{3}-67x^{2}+68x+672 }{ \color{blue}{ x + 3 } } = x^{3}-5x^{2}-52x+224 $$

Polynomial $ x^{3}-5x^{2}-52x+224 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{3}-5x^{2}-52x+224 $

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

Polynomial Equations Solver