Question

Letf(x)= 9x2+4 . (2x + 1)(x − 2)2
(i) Express f(x) in partial fractions.


(ii) Show that, when x is sufficiently small for x3 and higher powers to be neglected,
f(x) = 1 − x + 5x2.

What does it mean to show that x is sufficiently small for x^3 ?

Answer 1

is this supposed to be a fraction?

Answer 2

The first part yes, partial fraction, but I have the answers You want them?

Answer 3

i don't understand the expression
can you write it using the equation editor?

Answer 4

@satellite73
Let
\[f(x)=\frac{ 9x2+4 }{ (2x+1)(x−2)^2 }\]

i) Express f(x) in partial fractions.

(ii) Show that, when x is sufficiently small for x3 and higher powers to be neglected,
\[f(x)=1−x+5x^2.\]

Answer 5

lord is this ugly

Answer 6

i got the answer using wolfram
it is just a pain to do by hand
www.wolframalpha.com/input/?i=(9x^2%2B4)%2F((2x-1)(x-2)^2

Answer 7

I have the answer for a.

Answer 8

I just need to understand what they want me to do for b

Answer 9

Should I give you the anwers for a?

Answer 10

@satellite73 Answer for a.
\[\frac{ A }{ (2x+1) }+\frac{ B }{ (x – 2) }+\frac{ C }{ (x – 2)^2 }\]
Where A=1 B=4 C=8 such that
\[\frac{ 1 }{ (2x+1) }+\frac{ 4 }{ (x – 2) }+\frac{ 8 }{ (x – 2)^2 }\]

Answer 11

hold on that is not what i got

Answer 12

ooh you are right for sure sorry

Answer 13

for your second question i am at a loss
if you ignore the cube term in the denominator you get
\[\frac{9x^2+4}{-7x^2+4x+4}\]

Answer 14

maybe you are supposed to divide backwards

Answer 15

oh yeah, that works

Answer 16

use long division
\(4+4x-7x^2|4+0x+9x^2\)

Answer 17

it is very hard for me to write long division here, but that i think will give you what you want
set up the long division working from the bottom up instead of from the top down

Answer 18

But what does it mean to show that x is sufficiently small ?

Answer 19

it means "if x is sufficiently small ..." not "show that x is sufficiently small"

Answer 20

Don't worry I get it you expand binomially thanks for your help anyway