# Annoying exam question

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• December 30th 2008, 04:03 AM
Big_Joe
Annoying exam question
I think this is the right section for this question as i think you have to work with it algebraically. I'm really not sure what way to think about it. Could do with some help soon if possible:

Given that $\frac{2x^4-3x^2+x+1}{(x^2-1)}=(ax^2+bx+c)+\frac{dx+e}{(x^2-1)}$ find the values of the constants a, b, c, d and e. (4 marks)

Thanks in advance to those who help.
• December 30th 2008, 06:50 AM
Mush
Quote:

Originally Posted by Big_Joe
I think this is the right section for this question as i think you have to work with it algebraically. I'm really not sure what way to think about it. Could do with some help soon if possible:

Given that $\frac{2x^4-3x^2+x+1}{(x^2-1)}=(ax^2+bx+c)+\frac{dx+e}{(x^2-1)}$ find the values of the constants a, b, c, d and e. (4 marks)

Thanks in advance to those who help.

For $x \neq \pm 1$

Mutliply through by $x^2-1$:

$2x^4-3x^2+x+1 = (x^2-1)(ax^2+bx+c) + dx+e$

$2x^4 - 3x^2+x+1 = ax^4+bx^3+(c-a)x^2+(d-b)x+(e-c)$

Compare coefficients of $x^4$

$a = 2$

Compare coefficients of $x^3$

$b = 0$

Compare coefficients of $x^2$

$c-a = -3$

Compare coefficients of $x$

$d-b = 1$

Compare coefficients of $x^0$

$e-c= 1$
• December 30th 2008, 09:02 AM
Jester
Quote:

Originally Posted by Big_Joe
I think this is the right section for this question as i think you have to work with it algebraically. I'm really not sure what way to think about it. Could do with some help soon if possible:

Given that $\frac{2x^4-3x^2+x+1}{(x^2-1)}=(ax^2+bx+c)+\frac{dx+e}{(x^2-1)}$ find the values of the constants a, b, c, d and e. (4 marks)

Thanks in advance to those who help.

It might be a little easier (and quicker) if you do long division.
• December 30th 2008, 09:40 AM
Soroban
Hello, Big_Joe!

I don't suppose you tried long division . . . Naw!

Quote:

Given that: . $\frac{2x^4-3x^2+x+1}{x^2-1}\:=\:(ax^2+bx+c)+\frac{dx+e}{x^2-1}$

find the values of the constants $a, b, c, d, e.$

. . $\begin{array}{ccccccc}
& & & 2x^2 & & -1 \\
& & --- & --- & --- & --- \\
x^2-1 & ) & 2x^4 & -3x^2 & +x & +1 \\
& & 2x^4 & -2x^2 \\
& & --- & --- \\
& & & -x^2 & +x & +1 \\
& & & -x^2 && +1 \\
& & & --- & --- & --- \\
& & & & x
\end{array}$

Hence: . $\frac{2x^4 - 3x^2+x+1}{x^2-1} \:=\:2x^2-1 + \frac{x}{x^2-1}$

Therefore: . $a = 2,\;\;b = 0,\;\;c = -1,\;\;d=1,\;\;e=0$

• December 30th 2008, 03:08 PM
Big_Joe
Thanks! Both methods are useful to practice! I personally used long division method, but other is also pretty simple.