Improper partial fraction C4 question urgent help

q(x)=
8x^3+1
--------------
4x^2-4x+1

Show q(x) can be written
in the form Ax+B+ C/2x-1 + D/(2x-1)^2

I've got A as 2 by simply dividing the powers
but got 8 for B through algebraic division which is wrong
Don't know how to do C and D through partial fractions because of the denominator

(edited 5 years ago)

Reply 1

the bear

you can do long division ?

4x² - 4x + 1 ) 8x³ + 0x² + 0x + 1

Reply 2

mqb2766

Divide the numerator by the denominator before starting then do partial fractions on
remainder/denominator

Reply 3

Pakora99

Original post by the bear

you can do long division ?

4x² - 4x + 1 ) 8x³ + 0x² + 0x + 1

Original post by mqb2766

Divide the numerator by the denominator before starting then do partial fractions on
remainder/denominator

by doing algebraic long division I'm getting B as 8 which is wrong

Reply 4

mqb2766

Original post by Pakora99

by doing algebraic long division I'm getting B as 8 which is wrong

wanna show how?

Reply 5

Pakora99

Original post by mqb2766

wanna show how?

basically for a I got 2
so thats 2x on the top
then 2x(4x^2 -4x+1) = 8x^3 - 8x^2 +2x
then did
8x^3 + 0x^2 +0x +1
- (8x^3 -8x^2 +2x)
0--8= 8

(edited 5 years ago)

Reply 6

mqb2766

Original post by Pakora99

basically for a I got 2
so thats 2x on the top
then 2x(4x^2 -4x+1) = 8x^3 - 8x^2 +2x
then did
8x^3 + 0x^2 +0x +1
- (8x^3 -8x^2 +2x)
0--8= 8

The first term of the quotient is 2x because
(4x^2) * 2x = 8x^3
The second term of the quotient is "b" and you must have
2x*(-4x) + b*4x^2 = 0*x^2
so b = ?

Your quotient is (2x+b)

Some examples at
http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/long_division/long_division.html

(edited 5 years ago)

Reply 7

Pakora99

Original post by mqb2766

The first term of the quotient is 2x because
(4x^2) * 2x = 8x^3
The second term of the quotient is "b" and you must have
2x*(-4x) + b*4x^2 = 0*x^2
so b = ?

Your quotient is (2x+b)

so 8x^2 + (-)4x^2 (b) = 0x^2
so b = 2
ah I see what I did wrong thanks man

Reply 8

Pakora99

Original post by mqb2766

The first term of the quotient is 2x because
(4x^2) * 2x = 8x^3
The second term of the quotient is "b" and you must have
2x*(-4x) + b*4x^2 = 0*x^2
so b = ?

Your quotient is (2x+b)

for C and D how would I set this up because its
C
--------
(2x-1)

D
-------
(2x-1)^2

usually I'd do
C (2x-1) + D = 8x^3 +1
but Im not too sure

Reply 9

mqb2766

Original post by Pakora99

for C and D how would I set this up because its
C
--------
(2x-1)

D
-------
(2x-1)^2

usually I'd do
C (2x-1) + D = 8x^3 +1
but Im not too sure

You can say
(Cx + D)/(2x-1)^2
But your form is what you want.

You need to equate C & D to the remainder after division. You've already dealt with the quotient part.
remainder/denominator
is now a proper fraction and you do partial fractions on this. The original fraction was improper and you've transformed the problem into the
quotient + proper fraction

(edited 5 years ago)

Reply 10

Pakora99

Original post by mqb2766

where do I go from here tho after
Cx+D
--------
(2x-1)^2

Reply 11

mqb2766

Original post by Pakora99

where do I go from here tho after
Cx+D
--------
(2x-1)^2

What is the remainder after long polynomial division?

(8x^3+1)/(4x^2-4x+1) = quotient + remainder/(2x-1)^2

So quotient = Ax+B

and the partial fraction part C/2x-1 + D/(2x-1)^2 can be obtained from

remainder/(2x-1)^2

(edited 5 years ago)