# How do you solve 1/((x - 1)(x - 2)) + 1/((x - 2)(x - 3) )+ 1/((x - 3)(x - 4)) = 1/6?

May 8, 2016

$x = - 2$ or $x = 7$

#### Explanation:

Given:

$\frac{1}{\left(x - 1\right) \left(x - 2\right)} + \frac{1}{\left(x - 2\right) \left(x - 3\right)} + \frac{1}{\left(x - 3\right) \left(x - 4\right)} = \frac{1}{6}$

Multiply through by $6 \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right)$ to get:

$6 \left(x - 3\right) \left(x - 4\right) + 6 \left(x - 1\right) \left(x - 4\right) + 6 \left(x - 1\right) \left(x - 2\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right)$

Multiply out:

$6 \left({x}^{2} - 7 x + 12\right) + 6 \left({x}^{2} - 5 x + 4\right) + 6 \left({x}^{2} - 3 x + 2\right) = {x}^{4} - 10 {x}^{3} + 35 {x}^{2} - 50 x + 24$

$\rightarrow$

$18 {x}^{2} - 90 x + 108 = {x}^{4} - 10 {x}^{3} + 35 {x}^{2} - 50 x + 24$

Subtract the left hand side from the right to get:

${x}^{4} - 10 {x}^{3} + 17 {x}^{2} + 40 x - 84 = 0$

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 84$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 7$, $\pm 12$, $\pm 14$, $\pm 21$, $\pm 28$, $\pm 42$, $\pm 84$

Substituting $x = 2$ into the quartic we find:

${x}^{4} - 10 {x}^{3} + 17 {x}^{2} + 40 x - 84 = 16 - 80 + 68 + 80 - 84 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{4} - 10 {x}^{3} + 17 {x}^{2} + 40 x - 84 = \left(x - 2\right) \left({x}^{3} - 8 {x}^{2} + x + 42\right)$

Substituting $- 2$ into this cubic we find:

${x}^{3} - 8 {x}^{2} + x + 42 = - 8 - 32 - 2 + 42 = 0$

So $x = - 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{3} - 8 {x}^{2} + x + 42 = \left(x + 2\right) \left({x}^{2} - 10 x + 21\right)$

To factor and find the zeros of the remaining quadratic, note that $3 + 7 = 10$ and $3 \cdot 7 = 21$, so:

${x}^{2} - 10 x + 21 = \left(x - 3\right) \left(x - 7\right)$

So the remaining zeros are $x = 3$ and $x = 7$.

So all the zeros of our quartic polynomial are:

$- 2 , 2 , 3 , 7$

Note that the values $2$ and $3$ are not solutions of the original rational equation, since they result in zero denominators.

So the solutions of the original rational equation are $x = - 2$ and $x = 7$