Question

$$\frac{1}{5-t}+\frac{3}{5+t}+\frac{20}{25-t^2} = 0$$

Answer

$$ \begin{matrix}t_1 = 5 & t_2 = -5 & t_3 = 20 \end{matrix} $$

Explanation

$$ \begin{aligned} \frac{1}{5-t}+\frac{3}{5+t}+\frac{20}{25-t^2} &= 0&& \text{multiply ALL terms by } \color{blue}{ (5-t)\cdot(5+t)\cdot(25-t^2) }. \\[1 em](5-t)\cdot(5+t)\cdot(25-t^2)\cdot\frac{1}{5-t}+(5-t)\cdot(5+t)\cdot(25-t^2)\cdot\frac{3}{5+t}+(5-t)\cdot(5+t)\cdot(25-t^2)\cdot\frac{20}{25-t^2} &= (5-t)\cdot(5+t)\cdot(25-t^2)\cdot0&& \text{cancel out the denominators} \\[1 em]-t^3-5t^2+25t+125+3t^3-15t^2-75t+375-20t^2+500 &= 0&& \text{simplify left side} \\[1 em]2t^3-20t^2-50t+500-20t^2+500 &= 0&& \\[1 em]2t^3-40t^2-50t+1000 &= 0&& \\[1 em] \end{aligned} $$

$ \color{blue}{ 2x^{3}-40x^{2}-50x+1000 } $ 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 ( 2 ) are 1 2 .The factors of the constant term (1000) are 1 2 4 5 8 10 20 25 40 50 100 125 200 250 500 1000 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 10 }{ 2 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 20 }{ 2 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 25 }{ 2 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 40 }{ 2 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 50 }{ 2 } , ~ \pm \frac{ 100 }{ 1 } , ~ \pm \frac{ 100 }{ 2 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 125 }{ 2 } , ~ \pm \frac{ 200 }{ 1 } , ~ \pm \frac{ 200 }{ 2 } , ~ \pm \frac{ 250 }{ 1 } , ~ \pm \frac{ 250 }{ 2 } , ~ \pm \frac{ 500 }{ 1 } , ~ \pm \frac{ 500 }{ 2 } , ~ \pm \frac{ 1000 }{ 1 } , ~ \pm \frac{ 1000 }{ 2 } ~ $$

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(5) = 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 - 5} $

$$ \frac{ 2x^{3}-40x^{2}-50x+1000 }{ \color{blue}{ x - 5 } } = 2x^{2}-30x-200 $$

Polynomial $ 2x^{2}-30x-200 $ can be used to find the remaining roots.

$ \color{blue}{ 2x^{2}-30x-200 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

This page was created using

Equations Solver