$$\frac{4}{3(x+1)} = \frac{12}{(x+1)(x-1)}$$

Answer

$$ \begin{matrix}x_1 = 1 & x_2 = 2 & x_3 = 5 \end{matrix} $$

Explanation

$$ \begin{aligned} \frac{4}{3(x+1)} &= \frac{12}{(x+1)(x-1)}&& \text{multiply ALL terms by } \color{blue}{ 3(x+1)(x-1) }. \\[1 em]3(x+1)(x-1)\cdot\frac{4}{3(x+1)} &= 3(x+1)(x-1)\cdot\frac{12}{(x+1)(x-1)}&& \text{cancel out the denominators} \\[1 em]4x^3+4x^2-4x-4 &= 36x^2-72x+36&& \text{move all terms to the left hand side } \\[1 em]4x^3+4x^2-4x-4-36x^2+72x-36 &= 0&& \text{simplify left side} \\[1 em]4x^3-32x^2+68x-40 &= 0&& \\[1 em] \end{aligned} $$

$ \color{blue}{ 4x^{3}-32x^{2}+68x-40 } $ 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 ( 4 ) are 1 2 4 .The factors of the constant term (-40) are 1 2 4 5 8 10 20 40 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 4 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 10 }{ 2 } , ~ \pm \frac{ 10 }{ 4 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 20 }{ 2 } , ~ \pm \frac{ 20 }{ 4 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 40 }{ 2 } , ~ \pm \frac{ 40 }{ 4 } ~ $$

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} $

$$ \frac{ 4x^{3}-32x^{2}+68x-40 }{ \color{blue}{ x - 1 } } = 4x^{2}-28x+40 $$

Polynomial $ 4x^{2}-28x+40 $ can be used to find the remaining roots.

$ \color{blue}{ 4x^{2}-28x+40 } $ 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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