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by nickp33 » Fri Oct 10, 2008 1:21 pm
I have been prepairing for the GMAT for the past few months and I was working through practice problems in the GMAT Review 11th edition. Factoring is one of my weak points, so I try to concentrate on that area. Here is a practice problem I am having trouble with. I get it to the 3rd step and I can not in my life figure how they go from step 3 to 4. Once I'm at step 4, there is no problem, but I'm stuck between 3 and 4. Can someone please explain?
Thanks!

x^3-2x^2+x = -5(x-1)^2

Step 1: x^3-2x^2+x+5(x-1)^2=0
Step 2: x(x^2+2x+1)+5(x-1)2=0
Step 3: x(x-1)^2+5(x-1)^2=0
Here is where I get lost!
Step 4: (x+5)(x-1)^2=0
Step 5: X= -5 or X=1
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by cramya » Fri Oct 10, 2008 1:38 pm
Corrected step2 in bold and explained step4

Step 1: x^3-2x^2+x+5(x-1)^2=0

Step 2: x(x^2-2x+1)+5(x-1)^2=0

Step 3: x(x-1)^2+5(x-1)^2=0

Here is where I get lost!
Step 4: (x+5)(x-1)^2=0

Step 5: X= -5 or X=1

From step3 to step4 you can take the common factor (x-1) ^ 2
Therefore step4 is (x-1) ^ 2 (x+5)


Hope this helps!
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by vivek.kapoor83 » Sat Oct 11, 2008 12:14 am
taking factors..it pretty easy.pls use the same method as u take common in quad. equation
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by vishubn » Sat Oct 11, 2008 6:18 am
Factors !!
Step 1: x^3-2x^2+x+5(x-1)^2=0
Step 2: x(x^2+2x+1)+5(x-1)2=0
Step 3: x(x-1)^2+5(x-1)^2=0
Here is where I get lost!
Step 4: (x+5)(x-1)^2=0
Step 5: X= -5 or X=1
in step3 u came to the point where us see (x-1)^2 with different co efficients

so we know (x-1)^2 is common !

so (x-1)^2 (x+5) .... if u still confused ! ( multiply this line .. u get back to step 2)

Vishu
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