SOLUTION: Multiply (x+4)/(x^2+2x+1)*(x+1)/(x^2-16) (x^2-9)/(x^2+x-12)*(x^2-16)/(3) Please help me with this! Thank you!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Multiply (x+4)/(x^2+2x+1)*(x+1)/(x^2-16) (x^2-9)/(x^2+x-12)*(x^2-16)/(3) Please help me with this! Thank you!      Log On


   



Question 825738: Multiply
(x+4)/(x^2+2x+1)*(x+1)/(x^2-16)



(x^2-9)/(x^2+x-12)*(x^2-16)/(3)
Please help me with this! Thank you!

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%28%28x%2B4%29%2F%28x%5E2%2B2x%2B1%29%29%2A%28%28x%2B1%29%2F%28x%5E2-16%29%29
Way back in the "good ole days", when fractions had numbers for numerators and denominators, you should have learned that when multiplying fractions...
  • It is OK to cancel before multiplying. In fact this the only time when it is OK to cancel a factor in one fraction's numerator with a factor in another fraction's denominator. (This is called "cross-canceling" and is allowed only when multiplying fractions.)
  • Not only is it OK, it is a good idea. Canceling before multiplying makes the rest of the problem easier!
All of this is still true for these more complex fractions with variables in them. In fact, the advantage in canceling first is much greater with these fractions than it was with the simpler fractions!! So don't even think about multiplying first!

To cancel first we must know what the factors of each numerator and denominator are. So we start by factoring:

As we can see, there are many factor we can cancel:

leaving:
%281%2F%28x%2B1%29%29%2A%281%2F%28x-4%29%29
The advantage of canceling first should be obvious. These two fractions will be much easier to multiply than the original fractions! Multiplying we get:
1%2F%28x%5E2-3x-4%29

I'll leave the second problem for you to do. Be sure to factor and cancel first!

P.S. I think the second problem should be
%28%28x%5E2-9%29%2F%28x%5E2%2Bx-12%29%29%2A%28%28x%5E2-16%29%2F%28x-3%29%29
or
%28%28x%5E2-9%29%2F%28x%5E2%2Bx-12%29%29%2A%28%28x%5E2-16%29%2F%28x%2B3%29%29
and not what you posted:
%28%28x%5E2-9%29%2F%28x%5E2%2Bx-12%29%29%2A%28%28x%5E2-16%29%2F%283%29%29

P.P.S. This problem is not a rational function. It is a rational expression which should be posted in the "polynomials, rational expressions..." category.