# How do you solve (3x^2-10x-8)/(6x^2 +13x+6) *(4x^2-4x-15)/(2x^2-9x+10)=(x-4)/(x-2)?

May 21, 2015

If you multiply through by all the denominators (thank goodness for spreadsheets), you find:

$\left(3 {x}^{2} - 10 x - 8\right) \left(4 {x}^{2} - 4 x - 15\right) \left(x - 2\right)$
$= \left(12 {x}^{4} - 52 {x}^{3} - 37 {x}^{2} + 182 + 120\right) \left(x - 2\right)$
$= 12 {x}^{5} - 76 {x}^{4} + 67 {x}^{3} + 256 {x}^{2} - 244 - 240$

$\left(6 {x}^{2} + 13 x + 6\right) \left(2 {x}^{2} - 9 x + 10\right) \left(x - 4\right)$
$= \left(12 {x}^{4} - 28 {x}^{3} - 45 {x}^{2} + 76 x + 60\right) \left(x - 4\right)$
$= 12 {x}^{5} - 76 {x}^{4} + 67 {x}^{3} + 256 {x}^{2} - 244 - 240$

These will be equal for all values of $x \in \mathbb{R}$.

Consequently, the problem equation as stated is true for all $x \in \mathbb{R}$ except those values excluded because they cause one of the denominators to be zero.

$\Delta \left(6 {x}^{2} + 13 x + 6\right) = {13}^{2} - \left(4 \times 6 \times 6\right) = 169 - 144 = 25 = {5}^{2}$

So this quadratic factor has 2 rational zeros, viz

$x = \frac{- 13 \pm 5}{12}$ i.e. $- \frac{3}{2}$ and $- \frac{2}{3}$. These are excluded values.

$\Delta \left(2 {x}^{2} - 9 x + 10\right) = {9}^{2} - \left(4 \times 2 \times 10\right) = 81 - 80 = 1 = {1}^{2}$

So this quadratic factor has rational roots, viz

$x = \frac{9 \pm 1}{4}$ i.e. $2$ and $\frac{5}{2}$. These are excluded values.

The other excluded value would come from the factor $\left(x - 2\right)$, but that just gives us $2$ again.

So in summary, the given equation is true for all

$x \in \mathbb{R}$ \ $\left\{- \frac{3}{2} , - \frac{2}{3} , 2 , \frac{5}{2}\right\}$

May 21, 2015

Perhaps an easier way is to factor all of the quadratics first, so you can see all the linear factors...

$\left(3 {x}^{2} - 10 x - 8\right) = \left(3 x + 2\right) \left(x - 4\right)$
$\left(4 {x}^{2} - 4 x - 15\right) = \left(2 x - 5\right) \left(2 x + 3\right)$
$\left(6 {x}^{2} + 13 x + 6\right) = \left(3 x + 2\right) \left(2 x + 3\right)$
$\left(2 {x}^{2} - 9 x + 10\right) = \left(2 x - 5\right) \left(x - 2\right)$

These can be found using the standard solution for a quadratic:

$a {x}^{2} + b x + c = 0$ has discriminant $\Delta = {b}^{2} - 4 a c$ and
roots:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

For example, with $\left(3 {x}^{2} - 10 x - 8\right)$, we find roots $- \frac{2}{3}$ and $4$, which give us $\left(3 x + 2\right)$ and $\left(x - 4\right)$ as factors.

So we find:

$\frac{3 {x}^{2} - 10 x - 8}{6 {x}^{2} + 13 x + 6} \cdot \frac{4 {x}^{2} - 4 x - 15}{2 {x}^{2} - 9 x + 10}$

$= \frac{\left(3 x + 2\right) \left(x - 4\right)}{\left(3 x + 2\right) \left(2 x + 3\right)} \cdot \frac{\left(2 x - 5\right) \left(2 x + 3\right)}{\left(2 x - 5\right) \left(x - 2\right)}$

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

$= \frac{x - 4}{x - 2}$

Being identical to the right hand side, the original equation is satisfied for all $x \in \mathbb{R}$, except those that cause one of the denominators to become zero.

From our factorization into linear factors, we can quickly see that this happens when $x \in \left\{- \frac{3}{2} , - \frac{2}{3} , 2 , \frac{5}{2}\right\}$

So the equation is true for all $x \in \mathbb{R}$ \ $\left\{- \frac{3}{2} , - \frac{2}{3} , 2 , \frac{5}{2}\right\}$

May 21, 2015

Yet another approach is to eliminate one factor at a time, using synthetic division as follows:

Starting with:

$\frac{3 {x}^{2} - 10 x - 8}{6 {x}^{2} + 13 x + 6} \cdot \frac{4 {x}^{2} - 4 x - 15}{2 {x}^{2} - 9 x + 10} = \frac{x - 4}{x - 2}$

First note that $x = 2$ is excluded from being a solution, since it causes the denominator on the right to be 0. Remember this for later.

Now let us check to see if either of the numerators on the left hand side are divisible by $\left(x - 4\right)$. We can do this by substituting $x = 4$ into each of them in turn and seeing if the result is $0$. We can quickly discard $4 {x}^{2} - 4 x - 15$ as a possibility, since $15$ is odd and both of the other terms will be even for any $x \in \mathbb{Z}$. Let's try the other one:

$3 {x}^{2} - 10 x - 8 = 3 \times {4}^{2} - 10 \times 4 - 8 = 48 - 40 - 8 = 0$

So $\left(3 {x}^{2} - 10 x - 8\right)$ is divisible by $\left(x - 4\right)$ and we find:
$\left(3 {x}^{2} - 10 x - 8\right) = \left(x - 4\right) \left(3 x + 2\right)$.

Substituting this factorisation into our original equation we get:

$\frac{\left(x - 4\right) \left(3 x + 2\right)}{6 {x}^{2} + 13 x + 6} \cdot \frac{4 {x}^{2} - 4 x - 15}{2 {x}^{2} - 9 x + 10} = \frac{x - 4}{x - 2}$

We can now divide both sides of the equation by $\left(x - 4\right)$, eliminating the term #(x-4) to get:

$\frac{3 x + 2}{6 {x}^{2} + 13 x + 6} \cdot \frac{4 {x}^{2} - 4 x - 15}{2 {x}^{2} - 9 x + 10} = \frac{1}{x - 2}$

Notice that having eliminated one linear factor, we still have two left to try. Let's choose $\left(3 x + 2\right)$ next.

$\left(3 x + 2\right)$ cannot be a divisor of $\left(2 {x}^{2} - 9 x + 10\right)$ because the coefficient of ${x}^{2}$ is not divisible by $3$. So let's try the term $\left(6 {x}^{2} + 13 x + 6\right)$. Since this has symmetric coefficients, we quickly find that $\left(6 {x}^{2} + 13 x + 6\right) = \left(3 x + 2\right) \left(2 x + 3\right)$.

Substituting this in the equation, we get:

$\frac{3 x + 2}{\left(3 x + 2\right) \left(2 x + 3\right)} \cdot \frac{4 {x}^{2} - 4 x - 15}{2 {x}^{2} - 9 x + 10} = \frac{1}{x - 2}$

Notice this means that the values $x = - \frac{3}{2}$ and $x = - \frac{2}{3}$ are excluded as possible solutions, as they would cause the denominator of the first quotient term to be zero.

We can eliminate the common factor in the numerator and denominator of the first quotient to get:

$\frac{1}{2 x + 3} \cdot \frac{4 {x}^{2} - 4 x - 15}{2 {x}^{2} - 9 x + 10} = \frac{1}{x - 2}$

Repeating this process we eventually arrive at the equation:

$1 = 1$

...finding along the way that the value $x = \frac{5}{2}$ has to be excluded too. The advantage of this approach is that we do not have to use the general quadratic solution to find any of our factors.

So the original equation is true for all $x \in \mathbb{R}$ except those values of $x$ which would cause a denominator to be zero, viz the ones we found $\left\{- \frac{3}{2} , - \frac{2}{3} , 2 , \frac{5}{2}\right\}$.