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05-02-22 11:54:51 PM
Jul - General Chat - Math homework! New poll - New thread - New reply
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MaxKnight
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Posted on 10-01-07 06:54:16 AM Link | Quote
It's not mine, it's my brother's! He could do everything but problem 6, and I can't even remember how to do it right...

Anyways, here it is:

Problem 6. Polynomials, roots, and linear factors.
(A) If possible, produce a polynomial of degree 2 having 2, 6, and -13 as roots, and no other roots.
(B) If possible, produce two polynomials of degree 3 having 2, 6, and -13 as roots, and no other roots.
(C) If possible, produce a polynomial of degree 3 having 1089 as a y-indercept, and having 2, 6, and -13 as roots with no other roots.
(D) If possible. produce a polynomial of degree 4 having 2, 6, and -13 as roots, and no other roots.


Any ideas? This is College Algebra, which I'd've probably been able to solve if I was still in my High School Calculus course, but having been out of Math classes in general for about 4 years now, I have no idea how to solve these. He has to hand this work in in 5 and a half hours, and this is the only problem left. If anybody here knows anything about this, please speak up! (maybe NSNick has taken this class, since my brother is going to Ohio State too!)

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NSNick
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Posted on 10-01-07 11:24:29 AM Link | Quote




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I believe these should work:

A) Not possible.
B) 1/[(x-2)(x-6)(x+12)], putting any constant in the numerator works
C) 156816/[(x-2)(x-6)(x+12)]
D) 1/[(x-2)(x-6)(x+12)^2]

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MaxKnight
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Posted on 10-01-07 03:26:48 PM Link | Quote
Those worked with one exception: every (x+12) should be replaced with (x+13). My brother made that correction before turning it in, and found that it was correct.

Thanks for the help! Keeping my brother happy keeps me from being miserable!

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Skreeny
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Posted on 10-01-07 03:37:25 PM Link | Quote
...But aren't those rational functions rather than polynomials?

Perhaps college algebra is weird. I dunno.
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Posted on 10-01-07 03:40:26 PM Link | Quote
Originally posted by NSNick
I believe these should work:

A) Not possible.
B) 1/[(x-2)(x-6)(x+12)], putting any constant in the numerator works
C) 156816/[(x-2)(x-6)(x+12)]
D) 1/[(x-2)(x-6)(x+12)^2]



If you try to plug in the supposed roots into those functions you get undefined (division by zero). Correct me if I'm wrong, but roots are when a function equals zero.

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Posted on 10-01-07 03:56:28 PM Link | Quote
Yeah, all those fractions are upside-down. Make the constant be the denominator and they'll be correct, otherwise they're rational expressions and plugging the "roots" in will give division by zero, as Black Lord said.

Unless I completely misunderstood the question...
NSNick
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Posted on 10-02-07 03:21:54 AM Link | Quote




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Originally posted by Ran Yakumo
Yeah, all those fractions are upside-down. Make the constant be the denominator and they'll be correct, otherwise they're rational expressions and plugging the "roots" in will give division by zero, as Black Lord said.

Unless I completely misunderstood the question...

Oh yeah. Sorry, I'm stuck in my s-domain world. Whoops.

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Jul - General Chat - Math homework! New poll - New thread - New reply


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