a) how do I proceed from here in order to solve this?

b) I see (by plugging in) that the solutions -4 and 1 work. But shouldn’t I also look for a third solution, since the x is cubed? How is it possible to solve the same equation one way (the way you presented, which I completely understand) and get 2 values/solutions for x, and then solve it this way and get 3 values/solutions for x? Shouldn’t both ways end up with exactly the same solutions?
I must be making a very silly mistake but I can’t find it! Thank you.

Short answer is that in a multiple choice question, you don’t need to look for a third solution if none of the answer choices have three solutions. 🙂

Slightly longer answer: you can see by graphing (I wouldn’t recommend trying to factor a cubic expression) that there is a third solution to the equation you landed on: .

However, because the original equation has in a denominator, is extraneous. This is similar to solutions you might find after you square an equation that started with a —you have to go back and check them against the original equation.

## Comments (2)

Hi, quick question here:

what if I do not notice that x^2-x-6 can be replaced with (x-3)(x+2) and, instead, I cross multiply to solve. I get:

(2x+10)(x-3)=(x^2-x-6)(x+3)

2x^2-6x+10x-30= x^3+3x^2-x^2-3x-6x-18

2x^2+4x-30=x^3+2x^2-9x-18

x^3-13x+12=0

a) how do I proceed from here in order to solve this?

b) I see (by plugging in) that the solutions -4 and 1 work. But shouldn’t I also look for a third solution, since the x is cubed? How is it possible to solve the same equation one way (the way you presented, which I completely understand) and get 2 values/solutions for x, and then solve it this way and get 3 values/solutions for x? Shouldn’t both ways end up with exactly the same solutions?

I must be making a very silly mistake but I can’t find it! Thank you.

Short answer is that in a multiple choice question, you don’t need to look for a third solution if none of the answer choices have three solutions. 🙂

Slightly longer answer: you can see by graphing (I wouldn’t recommend trying to factor a cubic expression) that there is a third solution to the equation you landed on: .

However, because the

originalequation has in a denominator, is extraneous. This is similar to solutions you might find after you square an equation that started with a —you have to go back and check them against the original equation.