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Factoring as a potential method?

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by Baldini » Tue Mar 17, 2009 1:23 am
Hi,
came across this question:
If a^3 + a^2 - a - 1 = 0, then a =

a. 0
b. 1
c. 2
d. 3
e. 4

OA is B.

I know that one can find the answer out easily by plugging in the list of answers, but would it also be possible to find out the answer by factoring the equation, and how would one do it?

thanks in advance
GMAX
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by avenus » Tue Mar 17, 2009 1:43 am
In this case plugging the values in is easier and less time consuming, but if you're into factoring:

a^3 + a^2 - a - 1 = a^2(a - 1) + a^2 - 1 = a^2(a - 1) + (a + 1)(a - 1)=

= (a - 1)(a^2 + a + 1) = 0

Then
a = 1
or
(a^2 + a + 1) = 0, which has complex roots
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by Ian Stewart » Tue Mar 17, 2009 6:49 am
avenus wrote:In this case plugging the values in is easier and less time consuming, but if you're into factoring:

a^3 + a^2 - a - 1 = a^2(a - 1) + a^2 - 1 = a^2(a - 1) + (a + 1)(a - 1)=

= (a - 1)(a^2 + a + 1) = 0

Then
a = 1
or
(a^2 + a + 1) = 0, which has complex roots
Plugging in values is a perfectly good approach here. I think you flipped a sign or some other small error in the factorization above; it should be:

a^3 + a^2 - a - 1 = a^2(a+1) - (a+1) = (a^2 - 1)(a+1) = (a+1)(a-1)(a-1)

So there are two solutions to the equation: 1 and -1. The wording of the original question isn't good, since it implies there is only one solution.
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Re: Factoring as a potential method?

by Stuart@KaplanGMAT » Tue Mar 17, 2009 2:36 pm
Baldini wrote:Hi,
came across this question:
If a^3 + a^2 - a - 1 = 0, then a =

a. 0
b. 1
c. 2
d. 3
e. 4

OA is B.

I know that one can find the answer out easily by plugging in the list of answers, but would it also be possible to find out the answer by factoring the equation, and how would one do it?

thanks in advance
You can solve super quickly by knowing that the equation can be factored, without actually doing the factoring.

Let's just focus on the "-1" at the end of left side. If we're going to be multiplying integers (and looking at the rest of the equation and the answer choices, we're not worried about fractional solutions) to get a product of -1, we know that only +1 and -1 are potential answers.

Since +1 is among the choices and -1 isn't, choose (b)!

(Note: as Ian pointed out, the question is horribly worded; just by seeing -1 in the equation we know there will be both a positive and negative solution.)
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