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

##### 3 Answers

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

These will be equal for all values of

Consequently, the problem equation as stated is true for all

So this quadratic factor has 2 rational zeros, viz

So this quadratic factor has rational roots, viz

The other excluded value would come from the factor

So in summary, the given equation is true for all

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

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

roots:

For example, with

So we find:

Being identical to the right hand side, the original equation is satisfied for all

From our factorization into linear factors, we can quickly see that this happens when

So the equation is true for all

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

Starting with:

First note that

Now let us check to see if either of the numerators on the left hand side are divisible by

So

Substituting this factorisation into our original equation we get:

We can now divide both sides of the equation by

Notice that having eliminated one linear factor, we still have two left to try. Let's choose

Substituting this in the equation, we get:

Notice this means that the values

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

Repeating this process we eventually arrive at the equation:

...finding along the way that the value

So the original equation is true for all