How does one solve a polynomial absolute value inequality? say,
|(x-a)(x-b)(x-c)|>d

- inkyvoyd

How does one solve a polynomial absolute value inequality? say,
|(x-a)(x-b)(x-c)|>d

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- chestercat

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- inkyvoyd

No joke, I got a prob on a quiz for my AP BC calc class and I miserably got it wrong...

- inkyvoyd

the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.

- inkyvoyd

And, if I'm supposed to use a graphing calculator, tell me, because I honestly have no idea.

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## More answers

- phi

I think these are tricky.
I would re-write as
(x-a)(x-b)(x-c) - d > 0
or
-(x-a)(x-b)(x-c) - d > 0
multiply out and then factor to find the zeros
then look for intervals where the product of the factors are positive
very messy...

- inkyvoyd

Yes, but linear absolute value problems can have extraneous solutions...

- phi

then check the possible answers in the original equation

- inkyvoyd

So I probs just want a graphing calculator?

- phi

off hand, I don't know. do you have a specific example?

- inkyvoyd

Let me access my course. Sec

- inkyvoyd

|x^3-x^2+x-1|<3
"Round your answer to three decimal places and place in ascending order with a comma in between."
Answer: -0.811

- inkyvoyd

yeah, and my answer was from wolfram. I have no idea what I was supposed to do. and the lesson was not helpful at all. Didn't even mention it.

- phi

I'll think about it, as this is not something I've looked at very carefully.
But I not that x=1 gives 0 so 1 is part of the solution space.
Maybe by tomorrow I'll have something to post.

- inkyvoyd

Alright. Thanks for looking at it with me.

- amistre64

if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis

- amistre64

well, the curve does, then pick a point that is clearly not on the curve to test

- amistre64

|dw:1348175550151:dw|

- amistre64

|dw:1348175643801:dw|

- inkyvoyd

o.o

- inkyvoyd

it's a fake 6 degree polynomial!

- inkyvoyd

|dw:1348176111960:dw|

- inkyvoyd

Mmm.

- amistre64

its just the normal 3 degree with its negative parts popped up; since the absolute value of a number is never negative.

- inkyvoyd

Yeah. and then we draw that line. any way of an analytic solution, or are we just going to have to graph this guy?

- amistre64

im curious why the y=3 line

- amistre64

i see it now

- amistre64

yes, that y=3 line represents where the x axis would be if we subtract 3 from each side

- amistre64

|x^3-x^2+x-1| < 3
consider the values such that
(x^3-x^2+x-1) = 3
-(x^3-x^2+x-1) = 3

- amistre64

x^3-x^2+x-4 = 0
and
x^3-x^2+x+2 = 0
there is no good way to do a general cubic solution is there

- inkyvoyd

yeah I'm not cardano, I have 30 minutes to do this xD. I think it was a calculator prblem lol

- amistre64

do you need exact or approx ?

- inkyvoyd

approx.

- amistre64

x = 1.743 and -0.811 is what the google gives me as approx zeros

- amistre64

we can test those in the original format

- inkyvoyd

kay

- amistre64

at x=1.743; |x^3-x^2+x-1| -3 = 0.000xxx
at x=-0.811; |x^3-x^2+x-1| -3 = 0.00xxxx
and of course, when x=0, 1-3 is less than 0
so it should be an interval like: (-0.811,1.743) or however exact you need it to be

- inkyvoyd

I see. So that's what I'm supposed to do. Thank you.

- amistre64

http://www.wolframalpha.com/input/?i=%7Cx%5E3-x%5E2%2Bx-1%7C+-3%3C0
it even has the interval as a number line at the bottom

- phi

Here is what I came up with
|x^3-x^2+x-1|<3
we can do this with a graphing calculator
Case 1:
x^3-x^2+x-1 < 3
x^3-x^2+x-4 < 0
Examine the roots of x^3-x^2+x-4 = 0
Find 1 real root at x= 1.743 Graph of the function shows f(x)<0 for
x < 1.743
Case 2:
-(x^3-x^2+x-1) < 3
or, negating and flipping the sense of the relational operator
x^3-x^2+x-1 > -3
x^3-x^2+x+2 > 0
Find the roots of f(x)= x^3-x^2+x+2 =0
1 real root at x= -0.811
Graph shows f(x) > 0 for x > -0.811
Intersection of both regions
x < 1.743
x > -0.811
Gives -0.811 < x < 1.743

- inkyvoyd

I get why wolfram gave me the wrong answer. Cause it split it up into two inequalities. So mad lol.
and @phi , thanks for the answer :D

- phi

Sounds like you are ready for these... give 'em hell

- inkyvoyd

Too late :o
I got into BC calc 2.5 weeks late and I'm doing catch up work. that right there was week 1 or something, and I'm at week 2.5 tomorrow ;)
But thankfully I understand the concept, so I can take it on in event of unit test.

- inkyvoyd

It's not even calculus and I got it wrong. xS

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