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partial fractions

Struggling with this a level pure maths partial fractions question:
Express
(4y^2 + 3y 4) /y(2y −1)
in partial fractions.
Have tried my usual method for solving partial fractions but when I test by substituting in real values for y it doesn't work so I think theres a trick I'm missing. Would really appreciate the help! x
Reply 1
Original post by thatqueercoder
Struggling with this a level pure maths partial fractions question:
Express
(4y^2 + 3y 4) /y(2y −1)
in partial fractions.
Have tried my usual method for solving partial fractions but when I test by substituting in real values for y it doesn't work so I think theres a trick I'm missing. Would really appreciate the help! x

Both the numerator and the denominator are quadratics.
You need to divide first and get the linear remainder, then do partial fractions on

linear remainder/y(2y-1)

and don't forget to add in the quotient.
(edited 3 years ago)
Original post by mqb2766
Both the numerator and the denominator are quadratics.
You need to divide first and get the linear remainder, then do partial fractions on

linear remainder/y(2y-1)

and don't forget to add in the quotient.

That's great thank you... forgot that they can't have the same degree!
Original post by thatqueercoder
That's great thank you... forgot that they can't have the same degree!

Alternative approach is to recognise that, as the top and bottom are of the same degree, the answer will be in the form A + B/y + C/(2y - 1). Then you can proceed as usual to determine the values of A, B and C.

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