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 Post subject: Division of polinomials.Posted: Wed, 27 Jun 2012 08:39:57 UTC
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Joined: Sun, 25 Dec 2011 08:28:42 UTC
Posts: 12
Find the values of "a" and "b" if the
expression (x*x*x)+a(x*x)+bx-4 is exactly divisible by (x*x)-4.

Thanks for any possible solution because dividing such equations is extremely confusing to me.

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 Post subject: Re: Division of polinomials.Posted: Wed, 27 Jun 2012 13:45:09 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12075
Location: Austin, TX
thraiye wrote:
Find the values of "a" and "b" if the
expression (x*x*x)+a(x*x)+bx-4 is exactly divisible by (x*x)-4.

Thanks for any possible solution because dividing such equations is extremely confusing to me.

What equations? Neither of those has an equals sign...

Check out http://www.sosmath.com/algebra/factor/fac01/fac01.html for a nice primer, and also http://www.ebsinstitute.com/Baseball/EBS.crm5df2.html .

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 Post subject: Re: Division of polinomials.Posted: Wed, 27 Jun 2012 15:23:44 UTC
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Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7949
Location: Lexington, MA
Hello, thraiye!

Quote:

. .

If divides

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 Post subject: Re: Division of polinomials.Posted: Wed, 27 Jun 2012 15:30:05 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6005
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Soroban wrote:
Hello, thraiye!

Quote:

If divides

The constant term is -4, not +4.

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 Post subject: Re: Division of polinomials.Posted: Wed, 27 Jun 2012 16:49:02 UTC
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Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 3688
Location: Ottawa Ontario
If x=5, a=3, b=7 (as example):

X^3 + ax^2 + bx - 4 = 231
x^2 - 4 = 21
231 / 21 = 11 remainder 0

Soooo....what am I missing?

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 Post subject: Re: Division of polinomials.Posted: Wed, 27 Jun 2012 18:44:49 UTC
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Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7949
Location: Lexington, MA
For some reason, I'm not allowed to edit my post.
So, I'll correct my error in this new post.

Quote:

. .

If divides

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 Post subject: Division of polinomials.Posted: Fri, 29 Jun 2012 08:26:49 UTC
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Joined: Sun, 25 Dec 2011 08:28:42 UTC
Posts: 12
Thanks a lot everyone. [i][/i]

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