# Partial Fractions

• Jan 16th 2010, 12:38 PM
HNCMATHS
Partial 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 PM
e^(i*pi)
Quote:

Originally Posted by HNCMATHS
Hello,

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

$\displaystyle 6x^2 +5x -2 \div x(x-1)(2x+1)$

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

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

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

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 PM
Dinkydoe
For more examples of partial fractions:

Partial fraction - Wikipedia, the free encyclopedia
• Jan 16th 2010, 03:17 PM
Soroban
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: .$\displaystyle \frac{6x^2 +5x -2}{x(x-1)(2x+1)}$
Set up the equation: .$\displaystyle \frac{6x+5x-2}{x(x-1)(2x+1)} \;=\;\frac{A}{x} + \frac{B}{x-1} + \frac{C}{2x+1}$

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 PM
Plato
Quote:

Originally Posted by HNCMATHS
Could someone please help me by telling me what the best way to resolve the following equation into three partial fractions is?$\displaystyle 6x^2 +5x -2 \div x(x-1)(2x+1)$

Look at this result. It was done with a CAS costing about 40$US. There are even free programs, which will do this! The idea of doing partial fractions has gone the way of square roots and horseshoes. When has anyone done the square of 13 by hand? • Jan 17th 2010, 02:03 AM HNCMATHS Quote: Originally Posted by Archie Meade$\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)$and so on for B and C. *** Hello Archie, I understand what you have done up until the line starting... Now , you can ignore the denominators ?? Sorry :( Chris • Jan 17th 2010, 02:09 AM HNCMATHS My main problem is that i dont understand how to work out that values that we need to plug in for$\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$Its just clicked, again thank you all and i will be clicking thanks on all your posts. • Jan 17th 2010, 02:20 AM e^(i*pi) Quote: Originally Posted by HNCMATHS My main problem is that i dont understand how to work out that values that we need to plug in for$\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.