Marc S. answered • 01/02/13
Precalculus textbook editor can help you learn to find the answers
Let's start with the answer! Be sure to check out http://www.wolframalpha.com/input/?i=Simplify+%283q^2%2B9q%2B6%29%2F%289q^2-36q-45%29 . Wolfram Alpha today is like calculators were in the 1970s. Using it and playing with it will help you learn math. Now on to the skills you need to solve the problem.
I'm not sure what you mean by "removing factors of 1," but this expression can be simplified in multiple ways. The terms in the numerator have a common factor of 3:
3q2+9q+6 = 3(q2+3q+2)
In order to factor the trinomial q2+3q+2, we need to find two numbers with a sum of 3 and a product of 2. Those two numbers are 1 and 2:
3q2+9q+6 = 3(q2+3q+2) = 3(q+1)(q+2). Multiply (q+1)(q+2) by FOIL to check and see if q2+3q+2 returns. Factoring an expression into two parts is the opposite of multiplying two expressions together.
The terms in the denominator have a common factor of 9:
9q2−36q−45 = 9(q2−4q−5)
In order to factor the trinomial, q2−4q−5, we need to find two numbers with a sum of −4 and a product of −5. These two numbers are 1 and −5:
9q2−36q−45 = 9(q2−4q−5) = 9(q+1)(q−5). Again, try FOIL to check the product (q+1)(q−5)
After the numerator and denominator of your original expression are factored, we have:
(3q2+9q+6) / (9q2−36q−45) = 3(q+1)(q+2) / (9(q+1)(q−5))
Mulitplication and division can occur in any order, so 3/9 reduces to 1/3 and the q+1 on the top and bottom cancel completely to become 1. Now comes my chance to test the "strikethrough" font!
(3q2+9q+6) / (9q2−36q−45) = 3(q+1)(q+2) / ((3)9(q+1)(q−5)) = (q+2)/(3(q−5))
That's the form with a factored numerator and denominator, but I guess the simplest form is:
(q+2)/(3q-15)
which is why that Wolfram Alpha link above lists it first.
If you've learned about Domain and Range (these are usually pre-calculus topics) then a complete answer might indicate that q cannot equal −1 or 5 because either value would create a denominator of zero in the original function.
