SOLUTION: {{{((6x^2-5x-4)/(6x^2+5x+1))((3x^2+7x+2)/(3x^2-10x+8))}}}

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: {{{((6x^2-5x-4)/(6x^2+5x+1))((3x^2+7x+2)/(3x^2-10x+8))}}}      Log On


   



Question 623302:
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there-
.
The Problem: 


A Solution:
I take it that you want to simplify this expression as much as possible. It's usually the case in math
textbooks, that problems like these can be factored. As it turns out, all four of these trinomials are
factorable. 

We'll start with 6x^2 - 5x - 4. The first term, 6x^2, has factors 6x and x or 2x and 3x. The last term,
-4 has factors, 4 and -1, -4 and 1, or -2 and 2. We are looking for a combination of these factors with
a sum of the middle term, -5x. We see that 2x, 3x, 1, and -4 work because 
(3x)(1) + (2x)(-4) = 3x + -8x = -5x.

So, 
6x^2 - 5x - 4 = ( 2x + 1 )( 3x - 4 )

Using a similar procedure, we find that
6x^2 + 5x + 1 = ( 2x + 1 )( 3x + 1 )
3x^2 + 7x + 2 = ( 3x + 1 )( x + 2 )
3x^2 - 10x + 8 = ( x - 2 )( 3x - 4 )

Therefore,


Notice that the factors (2x+1), (3x-4), and (3x+1) are repeated in the numerator and the denominator.
There will "cancel out" because a number divided by itself equals one (2x+1)/(2x+1)=1, etc.) and
multiplying by one does not change the value of an expression.

Thus 


Here is a link to a polynomial calculator for finding factors. It includes good explanations:
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-factoring-calculator.php

If you find that you are confused about factoring with variables in general, here is a link to an
excellent tutorial:
http://www.purplemath.com/modules/simpfact.htm

Feel free to email if you have any questions about this solution.

Ms.Figgy
math.in.the.vortex@gmail.com