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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: IndefatigableLover on March 20, 2013, 11:49:30 pm

Title: Finding P(x)
Post by: IndefatigableLover on March 20, 2013, 11:49:30 pm: This question has lost me completely because the book doesn't explain anything nor can I find anything on the Internet on how to determine an equation with only one solution.

"Given that P(x) is a cubic polynomial with coefficient of $x^3$ being 1, and -1 is a solution of the equation P(x) = 0, find P(x)"
Title: Re: Finding P(x)
Post by: pi on March 20, 2013, 11:57:51 pm: Think y=(x-a)^3 form?
Title: Re: Finding P(x)
Post by: IndefatigableLover on March 21, 2013, 12:06:14 am: Quote from: pi on March 20, 2013, 11:57:51 pm
Think y=(x-a)^3 form?
I've tried that but the answer in the book differs from it :/

Answer:
Spoiler
x^3-2x+3
Title: Re: Finding P(x)
Post by: Professor Polonsky on March 21, 2013, 12:10:00 am: That answer has got to be wrong. If the coefficient of x^3 is 1, it can't just disappear.
Title: Re: Finding P(x)
Post by: IndefatigableLover on March 21, 2013, 12:16:09 am: Quote from: Polonius on March 21, 2013, 12:10:00 am
That answer has got to be wrong. If the coefficient of x^3 is 1, it can't just disappear.
Sorry about that... I accidentally imputed the wrong answer in but yeah that answer is in the book although 'pi's' method does work too so there might be multiple solutions for this questions >.<
Title: Re: Finding P(x)
Post by: Professor Polonsky on March 21, 2013, 12:24:59 am: I'm pretty sure there are infinite solutions for this question, though I might be wrong. Though the answer you provided still doesn't seem to be one of them...

Basically any $P(x)=x^3 + bx^2 + cx + d$ for which $P(-1)=0$ would be an answer.
Title: Re: Finding P(x)
Post by: IndefatigableLover on March 21, 2013, 12:30:12 am: Quote from: Polonius on March 21, 2013, 12:24:59 am
I'm pretty sure there are infinite solutions for this question, though I might be wrong. Though the answer you provided still doesn't seem to be one of them...

Basically any $P(x)=x^3 + bx^2 + cx + d$ for which $P(-1)=0$ would be an answer.
Yeah I was trying to work from the answer and backtrack (although I shouldn't) but I couldn't figure out how they did it...
I might just leave it with a statement saying there are infinite solutions for the question or something like that but thanks for your help Polonius :)
Title: Re: Finding P(x)
Post by: Professor Polonsky on March 21, 2013, 12:39:30 am: Which textbook is the question from, by the way? In any case, I wouldn't worry about this too much. The best you can do is provide a correct solution.
Title: Re: Finding P(x)
Post by: Mao on March 21, 2013, 10:34:03 am: There are two classes of solutions:

$P(x)=(x+1)(x-a)(x-b)$ for $a,b\in \mathbb{R}$ (note that it is possible for $a=b$ , and they can be negative)

$P(x)=(x+1-c^{1/3})^3+c$ for $c\in \mathbb{R}$ (note that $c$ can be negative)

So, any function that fits either of this form satisfies the original condition. There are infinite many solutions.