Question 623787
<pre>Hi, there---

Let's begin by trying to factor each of these trinomials. Typically, in textbook problems you 
will discover some common factors that you can then cancel out.

{{{x^2-9x+20}}}

We need two numbers whose product is 20 and whose sum is -9; the numbers are -4 and -5.
{{{x^2-9x+20=(x-4)(x-5)}}}

{{{2x^2-7x-15}}}
The leading coefficient is 2 so we have an extra step to find the factors. 2x^2 factors as (x)(2x) 
and -15 factors as (-3)(5) or (3)(-5). The numbers are -5 and 3 because (2x)(-5)-(x)(3)=-7x.
{{{2x^2-7x-15=(2x+3)(x-5)}}}

Using similar reasoning,
{{{2x^2-x-6=(2x+3)(x-2)}}}

{{{8+2x-x^2=(-1)(x^2-2x-8)=(-1)(x-4)(x+2)}}}

Therefore,
{{{((x^2-9x+20)/(2x^2-7x-15))((2x^2-x-6)/(8+2x-x^2))=((x-4)(x-5)*(2x+3)(x-2))/((2x+3)(x-2)(-1)(x-4)(x+2))}}}

Notice that (x-4), (2x+3), and (x-2) appear in the numerator and the denominator. We can 
"cancel these out" because a number divided by itself always equals 1. Any number multiply 
by 1 equals the same number.

Now we have 
{{{((x^2-9x+20)/(2x^2-7x-15))((2x^2-x-6)/(8+2x-x^2))=(x-5)/((-1)(x+2))}}}

Simplifying, we move the -1 outside the fraction.
{{{((x^2-9x+20)/(2x^2-7x-15))((2x^2-x-6)/(8+2x-x^2))=-((x-5)/(x+2))=(5-x)/(x+2)}}}

Please email me if you have questions about this solution. I'll be happy to help you sort it out.

Ms.Figgy
math.in.the.vortex@gmail.com
</pre>