Posted by Ragan on Friday, July 27, 2007 at 12:15am.
I am trying to factor polynomials by grouping the first two terms and the last two terms, but I am having some issues, I have done a few correct since I can look at the answers to the odd problems but it is taking me arund 25 minutes per problem. here is a sample problem in hope someone can explain it to be better than my book.
x to the 5th power + x cubed 2x squared2
x^5 + x^3  2x^2  2
x^3 is a common factor in the first two terms and the 2 is the only common factor in the last two.
so let's start with
x^3(x^2 + 1) ....
now look at the last two, they are both negative while the stuff at the front is positive. So the clue would be to factor out a 2
so
x^3(x^2 + 1)  2(x^2 + 1)
you now have two terms with x^2 + 1 as a new common factor
so to finish it off:
(x^2 + 1)(x^3  2) would be your final answer.
Notice that if we expand our answer, we get back the original expression, so our answer is correct.
These questions take practise and you might have to rearrange the terms into groups that belong together
ok, so if I had x^34x^2+2x8
in the first 2 terms x^34x^2; x^2 is my common factor and in the second set of terms 2x8, 2 is my common factor.
so it would look like this:
(x^2+4)(x2)
correct?
But what if I had this....2x10+xy5y?
in my first to terms 2x10 2 is the common factor right? and in the second y is my common factor correct?
so my answer would be...(52)(xy)?
I just noticed that was wrong.
it should be (x5)(2y).
Sorry about the mess up.
No.
for
x^34x^2+2x8
=x^2(x4) + 2(x4)
= (x  4)(x^2 + 2)
again no, and also no for your correction.
Here is the stepbystep
2x10+xy5y
= 2(x5) + y(x5) notice (x5) is the new common factor
=(x5)(2+y)
Are you not writing down the second lines in your solutions?
From that it would be obvious why it is +y and not y
Thanks, I am messing up my signs it looks like. But it looks like i have factored correctly. I will continue to practice, thanks for your help.

Algebra  dougie, Tuesday, January 29, 2008 at 3:20pm
123=3
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