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 Post subject: Division of polinomials.
PostPosted: Wed, 27 Jun 2012 08:39:57 UTC 
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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.
PostPosted: Wed, 27 Jun 2012 13:45:09 UTC 
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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.
PostPosted: Wed, 27 Jun 2012 15:23:44 UTC 
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Hello, thraiye!

Quote:
\text{Find the values of }a\text{ and }b\text{ if the  expression }P(x) \:=\:x^3+ax^2+bx-4
. . \text{is divisible by }x^2-4

If (x-2)(x+2) divides P(x),\,\text{ then }x = \pm2\text{ are zeros of }P(x).

\text{We have: }\:\begin{array}{ccccccc}x = 2: & 8 + 4a + 2b + 4 &=& 0 & [1] \\ x = \text{-}2: & \text{-}8 + 4a - 2b + 4 &=& 0 & [2] \end{array}

\begin{array}{cccccccccc}\text{Add [1] and [2]:} & 8a + 8 \:=\: 0 & \Rightarrow & \boxed{a \:=\: \text{-}1} \\ \\[-4mm] \text{Subtract [1] - [2]:} & 16 + 4b \:=\: 0 & \Rightarrow & \boxed{b \:=\: \text{-}4} \end{array}



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 Post subject: Re: Division of polinomials.
PostPosted: Wed, 27 Jun 2012 15:30:05 UTC 
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Soroban wrote:
Hello, thraiye!

Quote:
\text{Find the values of }a\text{ and }b\text{ if the  expression }P(x) \:=\:x^3+ax^2+bx-4
\text{is divisible by }x^2-4

If (x-2)(x+2) divides P(x),\,\text{ then }x = \pm2\text{ are zeros of }P(x).

\text{We have: }\:\begin{array}{ccccccc}x = 2: & 8 + 4a + 2b + 4 &=& 0 & [1] \\ x = \text{-}2: & \text{-}8 + 4a - 2b + 4 &=& 0 & [2] \end{array}

\begin{array}{cccccccccc}\text{Add [1] and [2]:} & 8a + 8 \:=\: 0 & \Rightarrow & \boxed{a \:=\: \text{-}1} \\ \\[-4mm] \text{Subtract [1] - [2]:} & 16 + 4b \:=\: 0 & \Rightarrow & \boxed{b \:=\: \text{-}4} \end{array}


The constant term is -4, not +4.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Division of polinomials.
PostPosted: Wed, 27 Jun 2012 16:49:02 UTC 
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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.
PostPosted: Wed, 27 Jun 2012 18:44:49 UTC 
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For some reason, I'm not allowed to edit my post.
So, I'll correct my error in this new post.


Quote:
\text{Find the values of }a\text{ and }b\text{ if the  expression }P(x) \:=\:x^3+ax^2+bx-4
. . \text{is divisible by }x^2-4

If (x-2)(x+2) divides P(x),\,\text{ then }x = \pm2\text{ are zeros of }P(x).

\text{We have: }\:\begin{array}{ccccccc}x = 2: & 8 + 4a + 2b - 4 &=& 0 & [1] \\ x = \text{-}2: & \text{-}8 + 4a - 2b - 4 &=& 0 & [2] \end{array}

\begin{array}{cccccccccc}\text{Add [1] and [2]:} & 8a - 8 \:=\: 0 & \Rightarrow & \boxed{a \:=\: 1} \\ \\[-4mm] \text{Subtract [1] - [2]:} & 16 + 4b \:=\: 0 & \Rightarrow & \boxed{b \:=\: \text{-}4} \end{array}



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 Post subject: Division of polinomials.
PostPosted: 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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