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For #1, use the formula: \(\sf a^2-b^2 = (a-b)(a+b)\)
I did number 1, but I am highly doubtful of my answer. Number 2 is confusing me a little.
What answer did you get?
I factored out the GCF and couldn't factor it out further, so I got: 9[9x^2 - 4y^2]
How did you even get 9?? Use the formula I gave you (:
I have no idea. I just used 9 because it was a common factor between 49 and 36. I will use your formula, thank you. :)
49 is a perfect square. Its only factors are: 1, 7, 49
Oh shoot careless mistake.
Thanks for pointing it out.
\(\sf a^2-b^2 = (a-b)(a+b)\) \(\sf 49x^2 - 36y^2 = (7x)^2 - (6y)^2\) Thus, implying that: \(\sf a \to 7x\) \(\sf b \to 6y\) Plug it into the formula :)
I think I understand the next problem now, I just factor it out (hopefully no careless mistakes this time!)
we're not done with #1 yet :p
(7x - 6y) (7x + 6y)?
Thanks for the help! This is the answer for #1?
For your second question, do you know how to group factor?
Yes, I think so.
and it is :)
Let me show you an example of group factoring \(\sf ax + bx + ay + by = x(a+b) + y(a+b) = (x+y)(a+b)\)
Oh yes, I do know how. Can you check my work/answers after I'm done?
sure :) I'd love to (:
x^3 + 2x^2 - 9x - 18 x^2(x + 2) -9(x + 2) (x + 2) (x^2 - 9) (x + 2) (x - 3) (x + 3)
Thank you so much for helping me and spending time to read this!!:D