$$\frac{7n}{3n+3}-\frac{5}{4n-4} = \frac{3n}{2n+2}$$

Answer

$$ \begin{matrix}n_1 = -1 & n_2 = -\dfrac{ 1 }{ 2 } & n_3 = 3 \end{matrix} $$

Explanation

$$ \begin{aligned} \frac{7n}{3n+3}-\frac{5}{4n-4} &= \frac{3n}{2n+2}&& \text{multiply ALL terms by } \color{blue}{ (3n+3)(4n-4)(2n+2) }. \\[1 em](3n+3)(4n-4)(2n+2)\frac{7n}{3n+3}-(3n+3)(4n-4)(2n+2)\cdot\frac{5}{4n-4} &= (3n+3)(4n-4)(2n+2)\frac{3n}{2n+2}&& \text{cancel out the denominators} \\[1 em]56n^3-56n-(30n^2+60n+30) &= 36n^3-36n&& \text{simplify left side} \\[1 em]56n^3-56n-30n^2-60n-30 &= 36n^3-36n&& \\[1 em]56n^3-30n^2-116n-30 &= 36n^3-36n&& \text{move all terms to the left hand side } \\[1 em]56n^3-30n^2-116n-30-36n^3+36n &= 0&& \text{simplify left side} \\[1 em]20n^3-30n^2-80n-30 &= 0&& \\[1 em] \end{aligned} $$

$ \color{blue}{ 20x^{3}-30x^{2}-80x-30 } $ 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 ( 20 ) are 1 2 4 5 10 20 .The factors of the constant term (-30) are 1 2 3 5 6 10 15 30 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 1 }{ 20 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 5 } , ~ \pm \frac{ 2 }{ 10 } , ~ \pm \frac{ 2 }{ 20 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 10 } , ~ \pm \frac{ 3 }{ 20 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 5 }{ 5 } , ~ \pm \frac{ 5 }{ 10 } , ~ \pm \frac{ 5 }{ 20 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 6 }{ 5 } , ~ \pm \frac{ 6 }{ 10 } , ~ \pm \frac{ 6 }{ 20 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 10 }{ 2 } , ~ \pm \frac{ 10 }{ 4 } , ~ \pm \frac{ 10 }{ 5 } , ~ \pm \frac{ 10 }{ 10 } , ~ \pm \frac{ 10 }{ 20 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 4 } , ~ \pm \frac{ 15 }{ 5 } , ~ \pm \frac{ 15 }{ 10 } , ~ \pm \frac{ 15 }{ 20 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 30 }{ 2 } , ~ \pm \frac{ 30 }{ 4 } , ~ \pm \frac{ 30 }{ 5 } , ~ \pm \frac{ 30 }{ 10 } , ~ \pm \frac{ 30 }{ 20 } ~ $$

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} $

$$ \frac{ 20x^{3}-30x^{2}-80x-30 }{ \color{blue}{ x + 1 } } = 20x^{2}-50x-30 $$

Polynomial $ 20x^{2}-50x-30 $ can be used to find the remaining roots.

$ \color{blue}{ 20x^{2}-50x-30 } $ 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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