$$7x^3-3x^2-35x+15 = 0$$

$$ \begin{matrix}x_1 = \dfrac{ 3 }{ 7 } & x_2 = -\sqrt{ 5 } & x_3 = \sqrt{ 5 } \end{matrix} $$

$ \color{blue}{ 7x^{3}-3x^{2}-35x+15 } $ 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 ( 7 ) are 1 7 .The factors of the constant term (15) are 1 3 5 15 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 7 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 7 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 7 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 7 } ~ $$

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 }{ 7 }) = 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}{ 7 x - 3 } $

$$ \frac{ 7x^{3}-3x^{2}-35x+15 }{ \color{blue}{ 7x - 3 } } = x^{2}-5 $$

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

$ \color{blue}{ x^{2}-5 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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