x3 - 7x + 6 = 0 has three solutions: x=1, 2, -3 /

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x^3 - 7x + 6 = 0 has three solutions: x=1, 2, -3

Can someone explain how to get these three values for x from this equation?

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by Mike@Magoosh » Fri Mar 30, 2012 2:23 pm
Hi, there. I'm happy to help with this. :)

I'll answer this question, but first let me make perfectly clear --- this is leaps and bounds beyond what the GMAT expects you to know.

Reasonably, the GMAT could give you f(x) = x^3 - 7x + 6, and ask you to plug in numbers --- which of the following makes f(x) = 0? --- and give you the numbers 1, 2, and -3 among others. BUT, as far as giving you the cubic function f(x) = x^3 - 7x + 6 and ask you, cold, to come up with the three roots --- without a calculator ---- that's viciously cruel, well beyond anything the GMAT would even think of asking.

We are getting into some advanced Precalculus topics here --- as with most of Precalculus, it's math beyond the scope of the GMAT. Everything in green below is math beyond the GMAT.

Inspecting f(x) = x^3 - 7x + 6, it's relatively easy to see that f(1) = 0. If x = 1 is root of f(x), that means (x - 1) is a factor of f(x). (That's the Factor Theorem.) This means, we can divide f(x) by (x - 1) to get a quadratic. We can accomplish that either by long division of polynomials or by synthetic division. I've attached an image of the successive steps of this synthetic division. Notice it ends with a remainder of zero, the guarantee that (x -1) went into f(x) evenly. The quotient polynomial is

q(x) = x^2 + x - 6

and this can be easily factored:

q(x) = x^2 + x - 6 = (x + 3)*(x - 2)

That immediately gives the other two roots, 2 & -3, so the three roots are {1, 2, -3}


To repeat, all of that is well beyond anything you would need to know for the GMAT. (Are you sorry you asked in the first place?)

Once again, the GMAT could reasonable give you the cubic and ask you to plug in numbers -- which of the following is a root --- that sort of thing. But actually factoring a cubic from scratch, without a calculator ---- that's leagues beyond what the GMAT can reasonably expect from test-takers.

I hope all this helps. Here's a much more GMAT like question:
https://gmat.magoosh.com/questions/114
When you submit your answer to that question, the following page will have the complete video solution. At Magoosh, we have 800+ GMAT questions, each with its own video solution. We also have 200+ lesson videos, covering all the content & strategy you will need for the GMAT. Of all the high quality GMAT prep available, Magoosh is the most affordable. We are currently having a sale that ends Tuesday 4/3, so now is a particularly opportune time to check us out.

Please let me know if you have any further questions, whether on the GMAT relevant stuff or on the Precalculus.

Mike :)
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by ShrutiN » Fri Mar 30, 2012 5:35 pm
If you ever see an equation with degree greater than 2, your first action should be hit and trial use common values of x like -2,-1,0,1,2. Most of the times one of these values will give you a solution, after that you can factorize the equation till it reduces to a quadratic and then solve it.

In the above equation; x^3-7x+6=0, intuitively x=1 is a solution
This means (x-1) is a factor of the equation, which can now be written as;

(x-1)(x^2+x-6)=0

The quadratic inside the bracket can be easily factorized to (x+3)(x-2)

Hence the equation becomes: (x-1)(x+3)(x-2)=0

This gives: x=1, x=-3 and x=2

Remember, there is no easy way to factorize equations higher than quadratic, you have to try the HIT and TRIAL method!

Hope that helps!!

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by angelgonzalez79 » Fri Mar 30, 2012 7:02 pm
Thanks guys! You are extremely helpful. I really appreciate the detailed feedback. I'm so happy this will not be on the GMAT!!!!!