Perceived Ambiguity in Factoring Polynomial Expressions

[xxpbdudexx]

As it turns out, this first part was an extremely pervasive user error that I have not seen for days. Still, though, it is interesting that equations can be factored in many different ways, so I will post that here instead:

3x^3 = -5x^2 +2x

-3x^3 -5x^2 + 2x = 0
Factored: x(-3x+1)(x+2) and x(3x-1)(-x-2)

x(3x^2 + 5x -2)
Factored: x(3x-1)(x+2) and x(-3x+1)(-x-2)

It seems as though any factoring of a trinomial can be made entirely of opposite signs and yet still produce the same solution sets.

Over the course of the last few days, I have found several equations which are so seemingly ambiguous that, when factored one way, produce one solution set, and, when factored another way, yield an entirely different solution set. In particular, I want to look at these two equations:

[STRIKE]3x^3 = -5x^2 +2x[/STRIKE]
abs(x^2 +10x) = 25
[STRIKE]
The first one was an example that my college algebra teacher used to explain factoring to the class.

I assumed that the easiest way to solve this equation was to simply subtract the 3x^3 to the other side of the equation. So then:

-3x^3 -5x^2 + 2x = 0

I then factored out an x:

x(-3x^2 -5x +2) = 0

Then finally, factoring the equation:

x(-3x+1)(x+2) = 0

which yields x = {0, -2, 1/3}

Note: Perhaps you already see what is going to happen. I know that factoring a -x out of the original equation would have yielded the same answer as the other equation, my concern is why it is necessary to do so.

My teacher, however, in her demonstration, decided to do it another way. She decided to move the -5x^2 and 2x to the other side with the 3x^3. So:

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

Here, I also noticed that the trinomial can be factored to both (3x-1)(x+2) and (-3x-1)(-x-2), but this is not so troubling since they both at least yield the same solution set.

The solution set here is

x = {0, -2, 1/3}

Simply put, how the hell does the same equation factored two different ways yield 1/3 and -1/3 but either way doesn't yield both?
[/STRIKE]
Then there is the second problem.

abs(x^2 +10x) = 25

I know that the equation requires one factoring and then one quadratic formula. Erm, to pose my exact question in a way that will probably make me seem rather stupid, why is doing this an invalid way to solve this equation?

x(x+10) = 25
x(x+10) = -25

x = {25, -25, 15, -35}

Or rather, what sort of rule/law/theorem/etc. is in place to prevent this kind of mathematical butchering from happening?

I think I'm going insane with these problems.

If this belongs in homework problems, sorry, but I didn't assume it did since I already know the answers to these problems and I am not merely asking for help.

 

[BloodyFrozen]

#1
-3x^3 -5x^2 + 2x = 0
Factored: x(-3x+1)(x+2) and x(3x-1)(-x-2)

All your doing is distributing the -1.

I'd proceed to solve it as:
-(3x³+5x²-2x)=0
-x(3x²+5x-2)=0
-x(3x-1)(x+2)=0

To get your first result, just multiply the -1 to (3x-1) and for the second result, just multiply the -1 to (x+2)

x(3x^2 + 5x -2)
Factored: x(3x-1)(x+2) and x(-3x-1)(-x+2) This is incorrect. Multiplying it out yields:
x(3x²-5x-2)

#2
Property

Zero product property

 

[xxpbdudexx]

#1
-3x^3 -5x^2 + 2x = 0
Factored: x(-3x+1)(x+2) and x(3x-1)(-x-2)

All your doing is distributing the -1.

I'd proceed to solve it as:
-(3x³+5x²-2x)=0
-x(3x²+5x-2)=0
-x(3x-1)(x+2)=0

To get your first result, just multiply the -1 to (3x-1) and for the second result, just multiply the -1 to (x+2)

x(3x^2 + 5x -2)
Factored: x(3x-1)(x+2) and x(-3x-1)(-x+2) This is incorrect. Multiplying it out yields:
x(3x²-5x-2)

#2
Property

Zero product property

I see. So the different factorings are merely due to a factored -1. That makes sense.

So then, with the second problem, the equation must be equal to zero in order to the zero product property? Okay, thanks, that helps a lot.

 

[BloodyFrozen]

So then, with the second problem, the equation must be equal to zero in order to the zero product property?

Yes

 

[AlephZero]

Then there is the second problem.

abs(x^2 +10x) = 25

I know that the equation requires one factoring and then one quadratic formula. Erm, to pose my exact question in a way that will probably make me seem rather stupid, why is doing this an invalid way to solve this equation?

x(x+10) = 25
x(x+10) = -25

x = {25, -25, 15, -35}

It is correct to say

"If x(x+10) = 0, then x = 0 or x+10 = 0, therefore x = 0 or x = -10"

But that is only valid if the right hand side is 0.

You can't say

"If x(x+10) = 25 then x = 25 or x+10 = 25"

That is completely wrong. You are trying to say something like
"25 x 1 = 25, therefore either x = 25 and x+10 = 1, or x = 1 and x+10 = 25"
which makes no sense.

The right way to solve this is to say

abs(x^2 +10x) = 25
Therefore either x^2 + 10x = 25, or x^2 + 10x = -25
Therefore either x^2 + 10x - 25 = 0, or x^2 + 10x + 25 = 0

And then solve the two quadratic equations separately.

One equation factorizes nicely, but you have to use the quadratic formula to solve the other one.

 

[Mark44]

So then, with the second problem, the equation must be equal to zero in order to the zero product property?

This is a fine point - an equation is not equal to any number. An equation is made up of two expressions that have the same value.

An equation can be either true or false. Some equations are always true; some are true only under certain conditions; and some are always false. Here are examples of all three.

x² + 2x + 1 = (x + 1)² (always true)
x² + 2x + 1 = 1 (true only for certain values of x: 0 or -2)
x + 1 = x + 3 (never true)