Express the following as a partial fraction:

a) f(x)=2x-13/(2x+1)(x-3)

b)f(x)=x^2 +5x +7/(x+2)^3

c) f(x)= x^2 -10/ (x-2)(x+1)

Sorry dont know how to do any of these.

Guest Nov 15, 2014

#7**+10 **

That just leaves one more

c)

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\

f(x)=\frac{x^2-10}{x^2-x-1}\\\\$$

We need to do an algebraic (or synthetic) division

$$\begin{tabular}{ccccccc} &&&\;1&&& &&&&||&||&||& \\

x^2&-x&-1&\|\;x^2&0&-10\\

&&&\;x^2&-x&-1\\ &&&||&||&||&& &&&&+x&-9&\\

\end{tabular}$$

so

$$\\\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{x^2-x-1}\\\\

\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{(x-2)(x+1)}\\$$

Now let

$$\\\frac{x-9}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\\\\

\frac{x-9}{(x-2)(x+1)}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\\\$$

Equating numerators we have

$$\\x-9=A(x+1)+B(x-2)\\\\

x-9=(A+B)x+(A-2B)\\\\

so\\

A+B=1\quad\rightarrow\quad A=1-B\\

-9=A-2B\\

-9=1-B-2B\\

-10=-3B\\

B=10/3\\\\

A=1-10/3\\

A=1-3/3-7/3\\

A=-7/3$$

therefore

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\

f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\

f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$

Melody Nov 16, 2014

#1**+5 **

a)

I haven't done these in ages. Let's see if I can remember.

There is probably a much simpler way of doing this but here goes

$$\\\frac{2x-13}{(2x+1)(x-3)}\\\\

=\frac{A}{2x+1}-\frac{B}{x-3}\\\\

=\frac{A(x-3)-B(2x+1)}{(2x+1)(x-3)}\\\\

=\frac{Ax-3A-2Bx-B}{(2x+1)(x-3)}\\\\

=\frac{(A-2B)x-(3A+B)}{(2x+1)(x-3)}\\\\

so\\\\

A-2B=2\qquad and \qquad 3A+B=13\\

A=2+2B\\

3(2+2B)+B=13\\

6+7B=13\\

7B=7\\

B=1\\

A=2+2*1\\

A=4\\

so\\\\

\frac{2x-13}{(2x+1)(x-3)}=\frac{4}{2x+1}-\frac{1}{x-3}$$

Melody Nov 15, 2014

#3**0 **

I am too tired to do more tonight - It is 1:30am.

The others are a little different I think. If you google it you are bound to find a good you tube that will show you how to do the others. There are lots of great maths you tubes and also other maths resource, that is how i often learn things. :)

Melody Nov 15, 2014

#5**+10 **

b)f(x)=x^2 +5x +7/(x+2)^3 = A/(x+2) + B/(x + 2)^2 + C/(x + 2)^3 ....multiply both sides by (x+2)^3....so we have

x^2 + 5x + 7 = A(x+2)^2 + B(x+2) + C

x^2 + 5x + 7 = A(x^2 + 4x + 4) +B(x + 2) + C

x^2 + 5x + 7 = A(x^2 + 4x + 4) +B(x + 2) + C

x^2 + 5x + 7 = Ax^2 +(4A+ B)x + (4A + 2B + C)

So, equating coefficients..

A = 1

4A + B = 5 → 4 + B = 5 → B = 1

4A + 2B + C = 7 → 4 + 2 + C = 7 → 6 + C = 7 → C = 1

So we have...

x^2 +5x +7/(x+2)^3 = 1/(x+2) + 1/(x + 2)^2 + 1/(x + 2)^3

CPhill Nov 16, 2014

#6**+5 **

Ah Chris you beat me to it I was just about to present my answer.

Luckily it is identical to your answer .

so Part B is now officially solved.

Melody Nov 16, 2014

#7**+10 **

Best Answer

That just leaves one more

c)

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\

f(x)=\frac{x^2-10}{x^2-x-1}\\\\$$

We need to do an algebraic (or synthetic) division

$$\begin{tabular}{ccccccc} &&&\;1&&& &&&&||&||&||& \\

x^2&-x&-1&\|\;x^2&0&-10\\

&&&\;x^2&-x&-1\\ &&&||&||&||&& &&&&+x&-9&\\

\end{tabular}$$

so

$$\\\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{x^2-x-1}\\\\

\frac{x^2-10}{(x-2)(x+1)}=1\;+\;\frac{x-9}{(x-2)(x+1)}\\$$

Now let

$$\\\frac{x-9}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}\\\\

\frac{x-9}{(x-2)(x+1)}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\\\$$

Equating numerators we have

$$\\x-9=A(x+1)+B(x-2)\\\\

x-9=(A+B)x+(A-2B)\\\\

so\\

A+B=1\quad\rightarrow\quad A=1-B\\

-9=A-2B\\

-9=1-B-2B\\

-10=-3B\\

B=10/3\\\\

A=1-10/3\\

A=1-3/3-7/3\\

A=-7/3$$

therefore

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\

f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\

f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$

Melody Nov 16, 2014