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Partial fractions

the question is, express x to the power of 2 divided by 4 - x to the power of 2 in partial fractions

I got the answer to be

1 divided by 2+x plus 1 divided by 2-x

the answer has a minus one on its own in it? I dont really get how
Original post by SunDun111
the question is, express x to the power of 2 divided by 4 - x to the power of 2 in partial fractions

I got the answer to be

1 divided by 2+x plus 1 divided by 2-x

the answer has a minus one on its own in it? I dont really get how


Remember that partial fractions can only be used when the degree of the numerator is less than the degree of the denominator. (The degree being the greatest power of x).
Is the question x^2/(4-x)^2 or x^2/(4-x^2)?
Original post by SunDun111
the question is, express x to the power of 2 divided by 4 - x to the power of 2 in partial fractions

I got the answer to be

1 divided by 2+x plus 1 divided by 2-x

the answer has a minus one on its own in it? I dont really get how


(4-x)^2 is not equal to (2+x)(2-x) unless you mean 4-x^2

If you have a squared bracket in the denominator you perform the brackets without the power in one fraction and then the entire thing as another. e.g. 2x/(x+1)^2 = A/(x+1) + B/(x+1)^2
(edited 7 years ago)
Reply 4
Original post by Math12345
Is the question x^2/(4-x)^2 or x^2/(4-x^2)?


the first one
Whichever one it is:

1. x24x2=(4x2)+44x2=1+44x2\frac{x^2}{4-x^2} = \frac{-(4-x^2)+4}{4-x^2} = -1 + \frac{4}{4-x^2}


2. (4x)2=x28x+16(4-x)^2=x^2-8x+16

x2(4x)2=(4x)2+8x16(4x)2=1+8x16(4x)2\frac{x^2}{(4-x)^2} = \frac{(4-x)^2+8x-16}{(4-x)^2} = 1 + \frac{8x-16}{(4-x)^2}

(You need to use long divsion basically)

Now use partial factions on the last fraction
(edited 7 years ago)
Reply 6
Original post by Vikingninja
(4-x)^2 is not equal to (2+x)(2-x) unless you mean 4-x^2

If you have a squared number in the denominator you perform the brackets without the power in one fraction and then the entire thing as another.


the power of 2, is inside, it is next to the x so its 4-x^2
Original post by SunDun111
the power of 2, is inside, it is next to the x so its 4-x^2


Original post by SunDun111
"Is the question x^2/(4-x)^2 or x^2/(4-x^2)?"
the first one


You said in this quote that its (4-x)^2


Original post by Math12345
naughty boy
Edit: One sec

It's against the rules to give answers.
(edited 7 years ago)
Reply 8
Original post by Math12345
Whichever one it is:

x24x2=(4x2)+44x2=1+44x2\frac{x^2}{4-x^2} = \frac{-(4-x^2)+4}{4-x^2} = -1 + \frac{4}{4-x^2}

Edit:

(4x)2=x28x+16(4-x)^2=x^2-8x+16

x2(4x)2=(4x)2+8x16(4x)2=1+8x16(4x)2\frac{x^2}{(4-x)^2} = \frac{(4-x)^2+8x-16}{(4-x)^2} = 1 + \frac{8x-16}{(4-x)^2}

(You need to use long divsion basically)

Now use partial factions on the last fraction


Original post by Vikingninja
You said in this quote that its (4-x)^2



It's against the rules to give answers.

crap I didnt mean to say the power of 2 was outside the bracket, it is inside, in the answer book my answer is right but im missing a minus one? I am wondering where it comes from?
Original post by SunDun111
crap I didnt mean to say the power of 2 was outside the bracket, it is inside, in the answer book my answer is right but im missing a minus one? I am wondering where it comes from?


Use long division or inspection like I did. You can't use partial fractions since the degree of the numerator and denominator is the same.
Reply 10
Original post by Math12345
Use long division or inspection like I did. You can't use partial fractions since the degree of the numerator and denominator is the same.


Ok thanks

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