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polak
Junior Member
Canada
5 Posts |
 Posted - 09/11/2009 : 19:58:01
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Hello,
I have this one question which I am having trouble to factor.
Equation:
a + b - c - d - 2ab + 2cd
Any help is appreciated, even helping me getting this started. Just don't know what to do.
Thank you in advance. |
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Mrspi
Advanced Member
USA
998 Posts |
Posted - 09/11/2009 : 20:28:30
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quote:
Originally posted by polak
Hello,
I have this one question which I am having trouble to factor.
Equation:
a + b - c - d - 2ab + 2cd
Any help is appreciated, even helping me getting this started. Just don't know what to do.
Thank you in advance.
When you're asked to factor any expression, the first thing you should look for is a factor that is common to all of the terms. In your problem, there is no common factor.
If the expression has more than three terms, you may want to consider "factoring by grouping." I think that's a possibility here.
Let's re-arrange the expression so that the terms involving "a" and "b" are together, and the terms involving "c" and "d" are together:
a - 2ab + b - c + 2cd - d
Ok...let's remove a common factor of -1 from the last three terms:
a - 2ab + b - 1(c - 2cd + d)
At this point, I'm going to turn it over to you.....
Please look at the first three terms...is this something you recognize? And look at the expression in the parentheses at "the end." Is this something you recognize?
Please tell us your thoughts at this point, so we can determine how best to help you.
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polak
Junior Member
Canada
5 Posts |
Posted - 09/11/2009 : 21:43:39
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quote:
Originally posted by Mrspi
[quote]Originally posted by polak
Let's re-arrange the expression so that the terms involving "a" and "b" are together, and the terms involving "c" and "d" are together:
a - 2ab + b - c + 2cd - d
Ok...let's remove a common factor of -1 from the last three terms:
a - 2ab + b - 1(c - 2cd + d)
At this point, I'm going to turn it over to you.....
Please look at the first three terms...is this something you recognize? And look at the expression in the parentheses at "the end." Is this something you recognize?
Please tell us your thoughts at this point, so we can determine how best to help you.
Thank you a bunch! Such a simple thing can alter an equation vastly. So basically after it was rearranged there were 2 Perfect Square Trinomials which could be factored, which I factored and was able to factor further with a Difference of Squares.
My work:
= a - 2ab + b - 1(c - 2cd + d)
= (a-b) - (c-d)
= (a-b+c-d)(a-b-c+d)
That's all that's to it. |
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polak
Junior Member
Canada
5 Posts |
Posted - 09/11/2009 : 21:56:52
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Hmm. I went through my whole worksheet and seems there are two that I can't figure out other than those.
The first is:
= 2x -3x+3x-2
What I thought I could do is:
= 2(x-1) + (3x-2) ... but that doesn't get anywhere.
Also tried:
= x(2x-3) + (3x-2) ... but I don't think I can get anywhere with that... It's close to Common Factoring but (2x-3) and (3x-2) are completely different.
The second is:
a+4t-4at+4t-2a
Not exactly sure where to go with this one. Should I be doing Common Factoring or Factoring a Trinomial.
Any help for these two is gladly appreciated.
Thank you! |
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Subhotosh Khan
Advanced Member
USA
9114 Posts |
Posted - 09/11/2009 : 23:15:18
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quote:
Originally posted by polak
Hmm. I went through my whole worksheet and seems there are two that I can't figure out other than those.
The first is:
= 2x-3x+3x-2
= 2x3 - 2 - 3x2 + 3x
= 2(x3 - 1) - 3(x -1)
= 2(x-1)(x2+ x + 1) - 3(x - 1)
Now continue
What I thought I could do is:
= 2(x-1) + (3x-2) ... but that doesn't get anywhere.
Also tried:
= x(2x-3) + (3x-2) ... but I don't think I can get anywhere with that... It's close to Common Factoring but (2x-3) and (3x-2) are completely different.
The second is:
a+4t-4at+4t-2a
=(a - 2t)2 - 2(a - 2t) now continue...
Not exactly sure where to go with this one. Should I be doing Common Factoring or Factoring a Trinomial.
Any help for these two is gladly appreciated.
Thank you!
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