Partial Fraction (P3)

The question is: (x^3+x^2-2x+4) / (x^2-4)

This is an improper fraction, so I did algebratic division first and got x+ {(2x)/(x^2-4)} as the answer.

Um, I got a different answer on this bit. I get 1 + x + (2x + 8)/(x^2 - 4)

Hence the complete partial fractions are 1 + x + 3/(x-2) - 1/(x+2)

That's the answer in the book, but the problem is I don't know how to get there. I already have the answer...

Bah, I'm still having trouble even with that correction.

Do you get x + (2x+4)/(x^2-4) when you do the algebratic division, because that's what I get.

Ah... I know you're right, but I can't do this algebratic division lol, I keep getting quotient as x and remainder as 2x + 4.

Don't worry, it's kinda hard to help online, I'll ask somebody in school today (gosh 1 am). Thanks for all ur help.

I'm doing P3, because I have 3 weeks of school after the exams. I just started P3.

ah right... lucky you

lou xxx

The year 12s go back to school and start P3, notice why its all algebra problems(first chapter). Cant wait until they move onto integration by substitution :P

....or some year 12s go back to school to start p4- notice that there aren't any recent p4 questions since the first chapter is inequalities

lol, theyve got to be difficult inequalities though surely?

no, not really. you just have to be more careful of one extra thing, e.g:

(x+1)/(x-1) > 1

you can't just multiply the (x-1) across, since you don't know the sign of the 'x' (if you mutliplied by a negative, then you would have to change the inequality...so since (x-1) could be negative, you can't just assume it would end up to be the same inequality- i.e. it may not still be "greater than"). so you have to take the '1' over:

(x+1)/(x-1) -1 > 0

now make it a single fraction:

2/(x-1) > 0

now you have to find the critcal values (when is the numerator 0, and when is the denominator 0). in this case, the numerator cannot be 0, so the only c.v. is when x=1

so then you look at the possible regions created by the critical value(s), i.e. as if drawn on a number line,
so x<1, you get a negative, and
x>1, you get a positive...so the answer is x>1

obviously in this case you don't need to do all the stuff with the critical values, since it can be done by inspection- but at least you can now know how to do the first chapter of p4!
i'll admit, though, that some of the ineqaulities you have ti solve by graphical methods can be a bit tricky.

The year 12s go back to school and start P3, notice why its all algebra problems(first chapter). Cant wait until they move onto integration by substitution :P

hehe yea. we're on to calculus with trig now which is ok at the mo- sure to get hard soon. although i find it hard to use the chain rule/product rule in my head all the time unlike some others in my class who can do it in seconds.
also those who have done the p3 exam, did u remember all those trig derivatives off by heart (eg cosecx), did u just derive them in the exam, or are they in the formulae booklet?