# Trick or Treat: What trick would you use in solving x^2 - x^3 = 12?

#### Steven G

##### Elite Member
What trick would you use in solving x2 -x3 = 12?

[imath]x^2-x^3-12[/imath] is factorable or are you looking for a clever "trick"?

What trick would you use in solving x2 -x3 = 12?
I see one obvious solution and from the graphs of x2 and x3 it's clear that it's the only one.

What trick would you use in solving x2 -x3 = 12?
Cardano's method. I'm just not that tricky.

-Dan

What trick would you use in solving x2 -x3 = 12?
I'd use the rational root theorem. That's very straightforward, but "tricks" don't have to be otherwise.

[imath]x^2-x^3-12[/imath] is factorable or are you looking for a clever "trick"?
Can you show how you'd factor it.

I guess that the rational root theorem is the best way to solve this.
I watch a video to by a Nigerian mathematician who solved it very differentially. For the record, the only reason why I mentioned that this professor is from Nigeria is because he was taught differentially and hence thinks differently those people from other areas.
He said to think of x2-x3= 12 as x2 - x3 - 22 - 23=0 and factor the lhs.
I guess this was overkill.

What trick would you use in solving x2 -x3 = 12?
I believe the method of observation is valid and that gives you x = -2

I believe the method of observation is valid and that gives you x = -2
Is -2 a triple root? If not, then you missed 2 roots.

How about complex roots?

Is -2 a triple root? If not, then you missed 2 roots.
If you plot f(x) = -x^3 + x^2 - 12, and put on your new glasses - you'll see that f(x) has only one real root.

Since we found one real root, we can invoke polynomial division ($$\displaystyle \frac{-x^3+x^2-12}{x+2}$$) and get a quadratic quotient that will show the nature of other two roots (no need to invoke tedious Cardano)

If you plot f(x) = -x^3 + x^2 - 12, and put on your new glasses - you'll see that f(x) has only one real root.

Since we found one real root, we can invoke polynomial division ($$\displaystyle \frac{-x^3+x^2-12}{x+2}$$) and get a quadratic quotient that will show the nature of other two roots (no need to invoke tedious Cardano)
But but but... I like tedious Cardano!

(Yes, I often do things the hard way.)

-Dan

I guess that the rational root theorem is the best way to solve this.
I watch a video to by a Nigerian mathematician who solved it very differentially. For the record, the only reason why I mentioned that this professor is from Nigeria is because he was taught differently and hence thinks differently those people from other areas.
He said to think of x2-x3= 12 as x2 - x3 - 22 - 23=0 and factor the lhs.
I guess this was overkill.
That last step is not immediately obvious to most students, though it isn't hard once you think about it. You end up with -(x+2)(x^2-3x+6), if I did that right, and you do get the complex roots as well. (The same is true of the rational root approach.)

How about complex roots?
The two missed roots I was referring to were complex roots.

That last step is not immediately obvious to most students, though it isn't hard once you think about it.
Possibly, in Nigeria the average student would think this way. I wonder. I enjoyed watching a number of youtube videos from this person, mathsonline tv, but he does do things differently--sometimes very differently.

Possibly, in Nigeria the average student would think this way. I wonder. I enjoyed watching a number of youtube videos from this person, mathsonline tv, but he does do things differently--sometimes very differently.
I was referring to the factoring step, which is peculiar because two terms are constants; it's the usual factoring by grouping, but with a twist, namely that one pair factor as a sum of cubes rather than just a GCF. (Or is there another method?)

The key step in the overall method is to see "by inspection" a useful decomposition of 12 (which is useful only because of the way you're going to be able to factor, which you don't know ahead of time!). The problem in teaching this as a method is that it's very specific to the numbers in the problem, and so not generalizable. Is any generalization offered?

On the other hand, "inspection" as in @lev888's solution (recognizing a solution) is a good general thing to learn, and amounts to a shortcut to the rational root method (trying a small list of possible solutions), just as immediately recognizing a pair of numbers to use in factoring a quadratic is a shortcut to a systematic search).

I too thought that his method was generalized. No generalization was offered.
He factored in a very unusual way--by always grouping.
For example, in factoring x2+ 5x + 6 he would say that you need to find two numbers that multiply out to 6 and add up to 5. The numbers he would say are 2 and 3. Then he would say x2+ 5x + 6 = (x2+ 2x) + (3x+ 6) and factor that--very bizarre in my opinion. In the end he gets the correct factorization.

He factored in a very unusual way--by always grouping.
For example, in factoring x2+ 5x + 6 he would say that you need to find two numbers that multiply out to 6 and add up to 5. The numbers he would say are 2 and 3. Then he would say x2+ 5x + 6 = (x2+ 2x) + (3x+ 6) and factor that--very bizarre in my opinion. In the end he gets the correct factorization.
This is just the standard "ac grouping" method that applies to any quadratic, applied to a case where its full power is not needed. I've told students that if they want to memorize one method, for all quadratics, this is fine, though often you'll say at the end, "I coulda just written (x+2)(x+3)!" Not bizarre, but overkill.

Did he factor x2 - x3 - 22 - 23 using a sum of cubes as I mentioned, or another way?

What trick would you use in solving x2 -x3 = 12?
[imath]\left\{ \begin{gathered} {x^2} - {x^3} = 12 \\ - {x^3} + {x^2} - 12 = 0 \\ {x^3} - {x^2} + 12 = 0 \\ \end{gathered} \right.[/imath] By inspection of the last, we see that [imath]x=-2[/imath] is a root.
Therefore, [imath](x+2)[/imath] must be a factor. So now we do long division on [imath]\large{\dfrac{{x^3} - {x^2} + 12}{(x+2)}}[/imath]
Doing so we get [imath]\large{x^3 - x^2+ 12=(x+2)(x^2-3x+6)}[/imath]
At this point we have reduced the remining parts to a good old quadratic: [imath]x^2-3x+6=0[/imath]
Apply the well known: [imath]\dfrac{b^2\pm\sqrt{b^2-4ac}}{2a}[/imath] so that [imath]a=1,~b=-3,~\&~c=6[/imath]
The discriminant: [imath]b^2-4ac=(-3)^2-4(1)(6)=-15[/imath] being negative tells us that there are two more roots, both complex roots.
[imath][/imath][imath][/imath][imath][/imath]

This is just the standard "ac grouping" method that applies to any quadratic, applied to a case where its full power is not needed. I've told students that if they want to memorize one method, for all quadratics, this is fine, though often you'll say at the end, "I coulda just written (x+2)(x+3)!" Not bizarre, but overkill.

Did he factor x2 - x3 - 22 - 23 using a sum of cubes as I mentioned, or another way?
He grouped it as (x2-22) - (x3 + 23)-----so he factored using the sum of cubes and the difference of squares.