The Student Room Group
Reply 1
Hi ViolinFirl.

Now that is very nearly correct, but theres one error in you method. Your dealing with a non-linear factor in the denominator ie the x^2 + 1, now when you have this in partial fractions you need to do something slightly different, like this:

1+2x(x2)(x2+1)=Ax2+Bx+cx2+1\displaystyle \frac{1 + 2x}{(x-2)(x^2 + 1)} = \frac{A}{x-2} + \frac{Bx + c}{x^2 + 1}

You see how there is a linear factor in the numerator of the second fraction on the RHS. hopefully this will help :biggrin:
You're trying to get your fraction into fractions of the form Ax2+1+Bx2\frac{A}{x^2 + 1} + \frac{B}{x-2}, right? But notice that x^2 + 1 has degree 2 (i.e. has an x^2 term in it), so you should actually be looking for something of the form Ax+Cx2+1\frac{Ax + C}{x^2 + 1} - in general, if the denominator has degree n, your numerator might well have anything up to degree n-1. If there was a cubic on the bottom, you'd need to put a quadratic on the top, etc.

Edit: snap.
Reply 3
ViolinGirl
(1+2x)/ (x-2)(x^2 + 1)

I've tried to break it up the normal way-->

1+2x = A/(x-2) + B(X^2 +1)

But on working backwards on the answer I get, I cannot get back to the original fraction, so I don't think this is correct.

Is there another method to do here?

Thanks.

Unparseable latex formula:

\displaystyle\[br]1+2x\equiv\frac{A}{x-2}+\frac{Bx+C}{x^2+1}



edit:uughh badly beaten:sigh:

(1+2x)/ (x-2)(x^2 + 1)

I've tried to break it up the normal way-->

1+2x = A/(x-2) + B(X^2 +1)

But on working backwards on the answer I get, I cannot get back to the original fraction, so I don't think this is correct.

Is there another method to do here?

Thanks.


The expression you want is

Ax2+Bx+Cx2+1 \frac{A}{x-2} + \frac{Bx+C}{x^2+1}

This is because the x^2 term means there is a quadratic factor in the denominator, so you need more than just B.

Check here (bottom of page) for a better explanation than mine!

http://www.mathsrevision.net/alevel/pages.php?page=94
Reply 5
Thanks to you both! :smile: