Hello,

Could someone please help me by telling me what the best way to resolve the following equation into three partial fractions is?

$\displaystyle 6x^2 +5x -2

\div

x(x-1)(2x+1)$

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- Jan 16th 2010, 12:38 PMHNCMATHSPartial Fractions
Hello,

Could someone please help me by telling me what the best way to resolve the following equation into three partial fractions is?

$\displaystyle 6x^2 +5x -2

\div

x(x-1)(2x+1)$ - Jan 16th 2010, 12:44 PMe^(i*pi)
$\displaystyle \frac{6x^2+5x-2)}{x(x-1)(2x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{2x-1}$

Which becomes $\displaystyle 6x^2+5x-2 = A(x-1)(2x+1) + B(x)(2x+1) + C(x)(x-1)$

It's best to find $\displaystyle f(0)$, $\displaystyle f(1)$ and $\displaystyle f\left(-\frac{1}{2}\right)

$ - Jan 16th 2010, 12:48 PMHNCMATHS
Sorry I do not understand your reply

- Jan 16th 2010, 01:46 PMArchie Meade
$\displaystyle \frac{A}{x}+\frac{B}{x-1}+\frac{C}{2x+1}=\frac{A(x-1)(2x+1)+Bx(2x+1)+Cx(x-1)}{x(x-1)(2x+1)}$

$\displaystyle =\frac{6x^2+5x-2}{x(x-1)(2x+1)}$

Now, you can ignore the denominators and multiply out the numerator on the right and compare terms with the left to find A, B and C.

The reason $\displaystyle e^{i*pi}$ recommended now finding f(0), f(1) and f(-0.5) is because f(0) finds A since the parts with B and C will be zero.

You get $\displaystyle 6(0)+5(0)-2=A(-1)(1)$ and so on for B and C. - Jan 16th 2010, 01:47 PMDinkydoe
For more examples of partial fractions:

Partial fraction - Wikipedia, the free encyclopedia - Jan 16th 2010, 03:17 PMSoroban
Helolo, HNCMATHS!

I'll show you the method I was taught.

I think it's much simpler; don't know why everyone doesn't teach it.

Quote:

Resolve into partial fractions: .$\displaystyle \frac{6x^2 +5x -2}{x(x-1)(2x+1)}$

Multiply by the LCD: .$\displaystyle 6x^2 + 5x - 2 \;=\;(x-1)(2x+1)A + x(2x+1)B + x(x-1)C $

Select some "good" values for $\displaystyle x\!:$

Let $\displaystyle x=0:$ . . . . . $\displaystyle 0 + 0 - 2 \:=\:(\text{-}1)(1)A + 0\!\cdot\! B + 0\!\cdot\! C$ . . . . . $\displaystyle \Rightarrow$ . . . $\displaystyle \text{-}A \:=\:\text{-}2\quad\Rightarrow\quad\;\,\boxed{ A \:=\:2}$

Let $\displaystyle x = 1:$ . . . . $\displaystyle 6+5-2 \:=\:0\cdot A + (1)(3)B + 0\cdot C$ . . . . . .$\displaystyle \Rightarrow\qquad 3B \:=\: 9 $ . . $\displaystyle \Rightarrow\quad\;\boxed{ B\: =\: 3}$

Let $\displaystyle x = \text{-}\tfrac{1}{2}:\;\;6\left(\tfrac{1}{4}\right) + 5\left(\text{-}\tfrac{1}{2}\right)-2 \:=\: 0\cdot A + 0\cdot B + \left(\text{-}\tfrac{1}{2}\right)\left(\text{-}\tfrac{3}{2}\right)C \quad\Rightarrow$ . . .$\displaystyle \tfrac{3}{4}C \:=\:\text{-}3 \quad\;\;\Rightarrow\quad\boxed{ C \:=\:\text{-}4}$

Therefore: .$\displaystyle \frac{6x^2+5x-2}{x(x-1)(2x+1)} \;=\;\frac{2}{x} + \frac{3}{x-1} - \frac{4}{2x+1} $

[Edit: oops! .I see that $\displaystyle e^{i\pi}$ recommended this method.]

- Jan 16th 2010, 03:45 PMPlato
- Jan 17th 2010, 02:03 AMHNCMATHS
- Jan 17th 2010, 02:09 AMHNCMATHS
My main problem is that i dont understand how to work out that values that we need to plug in for $\displaystyle x$

EDIT**

I think I have worked it out, $\displaystyle x$ has to equal 0 in the divisor? so in our case where we have $\displaystyle 2x+1$ ..... $\displaystyle x$ would equal $\displaystyle -0.5$

Its just clicked, again thank you all and i will be clicking thanks on all your posts. - Jan 17th 2010, 02:20 AMe^(i*pi)