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Help with an Algebra PS question

by r321 » Fri Nov 12, 2010 3:19 am
If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve this equation through algebra as well.

Thanks!
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by beat_gmat_09 » Fri Nov 12, 2010 3:42 am
r321 wrote:If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve this equation through algebra as well.

Thanks!
Denominator is x-5, try to factor numerator by rearranging/rewriting
Rewrite the numerator - x^3 - 5x^2 - x^2 - x + 30
x^2(x - 5) - (x^2 + x -30)
x^2(x - 5) - (x + 6) (x - 5)
(x-5) [x^2 - (x + 6)]
(x-5) (x^2 - x - 6)
Dividing by (x - 5) = x^2 - x - 6
Option A.
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by r321 » Fri Nov 12, 2010 4:10 am
beat_gmat_09 wrote:
r321 wrote:If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve this equation through algebra as well.

Thanks!
Denominator is x-5, try to factor numerator by rearranging/rewriting
Rewrite the numerator - x^3 - 5x^2 - x^2 - x + 30
x^2(x - 5) - (x^2 + x -30)
x^2(x - 5) - (x + 6) (x - 5)
(x-5) [x^2 - (x + 6)]
(x-5) (x^2 - x - 6)
Dividing by (x - 5) = x^2 - x - 6
Option A.
Thank you!! I knew factoring out (x-5) from the numerator was the key but couldn't figure out how to do that...
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by Rahul@gurome » Fri Nov 12, 2010 9:36 pm
There is a shorter way of solving.
If H = [(x^3) - 6(x^2) - x + 30] / (x-5), we get that H * (x - 5) = (x^3) - 6(x^2) - x + 30.
Now the constant term on the left hand side = the constant term on right hand side.
So (constant term of H) * (-5) = 30.
So constant term of H is -6.
Check the options now.
Only A has the constant term as -6.
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