Factoring inequalities?

[yo4561]

Good morning MP!

There was a question emailed out by my instructor as a drill exercise, and I struggled in dealing with this inequality:

v^4 <v

I initially moved the v to the left and did --> v(v^3-1)<0 --> then I believe that I did an illegal math move of v<0 and v^3<1.

When is appropriate to factor inequalities and how would you break them up if you can to solve for the answer? If an example could be provided, I would be so appreciative.

Thank you MP you are the BEST! Don't forget it!

[esledge]

v^4 <v

I initially moved the v to the left and did --> v(v^3-1)<0 --> then I believe that I did an illegal math move of v<0 and v^3<1.

When is appropriate to factor inequalities and how would you break them up if you can to solve for the answer? If an example could be provided, I would be so appreciative.

Good job NOT just dividing by v initially! (For anyone else reading this, that would be dangerous: what if v is zero? what if v is negative?)

Pretend the inequality is an equal sign, so v(v^3 - 1) = 0, and that does imply that v = 0 or v = 1, which is what you were wanting to do next. Sketch out a number line, mark v = 0 and v = 1 (splitting the number line into three sections), and then plug a test value from each of the three "sections" of the number line into the original inequality:

If v = -2, then is v^4 < v? Is +16 < -2? No, so this rules out the entire v < 0 range.
If v = 1/2, then is v^4 < v? Is 1/16 < 1/2? Yes, so this indicates that the entire 0 < v < 1 range works.
If v = 3, then is v^4 < v? Is 81 < 3? No, so this rules out the entire v > 1 range.

The resulting solution is 0 < v < 1.

Thank you MP you are the BEST! Don't forget it!

Thank you!

[yo4561]

Thank you so much Emily! To gently confirm, is there any situation in which you would want to factor an inequality?

[esledge]

In the solution I suggested, note that you still do factor to the point where you get v(v^3-1)<0. I didn't mean to suggest that factoring is bad! Indeed, we used the factored form to decide that v=0 and v=1 were the "limits" of the potential solutions, so to speak. From there, I just suggested plugging test cases to decide which ranges are in or out, rather than continuing algebra steps.

The alternative is to sketch all of this on a number line and label the number line with the signs of the factored terms.

For v < 0, the v term is negative and the (v^3-1) term is negative. The product is positive, so this range is NOT a solution to v(v^3-1)<0.
For 0 < v < 1, the v term is positive and the (v^3-1) term is negative. The product is negative, so this range IS a solution to v(v^3-1)<0.
For v > 1, the v term is positive and the (v^3-1) term is positive. The product is positive, so this range is NOT a solution to v(v^3-1)<0.

So, see how we still factored in both of these solutions? The difference is just what you do next: plug specific test numbers or think of number properties (signs). But neither approach used algebra past the factoring step; that's where your mistakes were made!