SOLUTION: Hello, I'm having trouble trying to factor each polynomial completely. The equation is 4x^2-9y^2 In words the equation would be four x squared minus nine y squared Thank you so m

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello, I'm having trouble trying to factor each polynomial completely. The equation is 4x^2-9y^2 In words the equation would be four x squared minus nine y squared Thank you so m      Log On


   



Question 484407: Hello, I'm having trouble trying to factor each polynomial completely. The equation is 4x^2-9y^2
In words the equation would be four x squared minus nine y squared
Thank you so much for the help.
-Kaitlin

Answer by tinbar(133) About Me  (Show Source):
You can put this solution on YOUR website!
When trying to factor polynomials, there are some special 'forms' that polynomials can be given in which result is quick and easy factoring.
Your questions happens to be in that form; namely in the form a^2 - b^2, where a and b are some terms, they can be numbers, variables, or a mix of both. In this case, they are a mix of both. Anyway, if we can figure out what our a and b are we can quickly factor this since a^2 - b^2 = (a-b)*(a+b) [Take these bracket and re-expand them to make sure, don't just take my word!]
Now in your question we can see the first term is 4x^2. So some a^2 = 4x^2. If we want a final result of the x^2 term before even considering the 4 in front of it, we can agree that our 'a' term must have an x in it, since when we square a, we will square x and get x^2, but we are missing the 4, so what else do we need in the 'a' term so that when we square it we get 4? Well, we know 2^2 is 4. So our 'a' is really just 2x, since a^2 is 4x^2. Similarly, we find that our 'b' should be 3y.
Our final answer is: 4x^2-9y^2 = (2x-3y)*(2x+3y)[re-expand these two brackets to make sure they multiply out into the original equation]