Partial Fractions

**racheltllong** · June 4th 2008, 02:13 AM

Hi there
Please could someone help me to solve this question?

Express ((8x)^3 - (18x)^2) / (x-1)(4x+1)

in the form Ax + B + (C/(x-1)) + (D/(4x+1))

Thank you all for your help!
Rachel

**Soroban** · June 4th 2008, 03:01 AM

Hello. Rachel!

Express . $\frac{(8x)^3 - (18x)^2}{(x-1)(4x+1)}$ . in the form: . $Ax + B + \frac{C}{x-1} + \frac{D}{4x+1}$

Using long division, we have: . $\frac{512x^3 - 324x^2}{4x^2 - 3x - 1} \;=\;128x + 15 + \frac{173x + 15}{(x-1)(4x+1)}$

Then: . $\frac{173x + 15}{(x-1)(4x+1)} \;=\;\frac{C}{x-1} + \frac{D}{4x+1} \quad\Rightarrow\quad 173x + 15 \;=\;(4x+1)C + (x-1)D$

Let $x = 1\!:\;\;188 \:=\:5C\quad\Rightarrow\quad C \:=\:\frac{188}{5}$

Let $x = -\frac{1}{4}\!:\;\;-\frac{173}{4} + 15 \:=\:-\frac{5}{4}D \quad\Rightarrow\quad D \:=\:\frac{113}{5}$

Therefore: . $\frac{512x^3 - 324x^2}{(x-1)(4x+1)} \;=\;128x + 15 + \frac{\frac{188}{5}}{x-1} + \frac{\frac{113}{5}}{4x+1}$

Thread: Partial Fractions

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