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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.

 Nov 15, 2014

Best Answer 

 #7
avatar+105604 
+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

 

$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\
f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$

.
 Nov 16, 2014
 #1
avatar+105604 
+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

$$\\\frac{2x-13}{(2x+1)(x-3)}\\\\
=\frac{A}{2x+1}-\frac{B}{x-3}\\\\
=\frac{A(x-3)-B(2x+1)}{(2x+1)(x-3)}\\\\
=\frac{Ax-3A-2Bx-B}{(2x+1)(x-3)}\\\\
=\frac{(A-2B)x-(3A+B)}{(2x+1)(x-3)}\\\\
so\\\\
A-2B=2\qquad and \qquad 3A+B=13\\
A=2+2B\\
3(2+2B)+B=13\\
6+7B=13\\
7B=7\\
B=1\\
A=2+2*1\\
A=4\\
so\\\\
\frac{2x-13}{(2x+1)(x-3)}=\frac{4}{2x+1}-\frac{1}{x-3}$$

.
 Nov 15, 2014
 #2
avatar
0

ahh thank you! that's perfect

 Nov 15, 2014
 #3
avatar+105604 
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. :)

 Nov 15, 2014
 #4
avatar
0

Oh ahaha it's 14.51 here! Okay, I'll look into it now. Thankyou!

 Nov 15, 2014
 #5
avatar+104793 
+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

 

 Nov 16, 2014
 #6
avatar+105604 
+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.    

 Nov 16, 2014
 #7
avatar+105604 
+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
 #8
avatar+104793 
0

Very nice, Melody.....I didn't know what to do when the powers on each polynomial in both numerator and denominator were equal....

Thanks for that "refresher"....(although.....if I didn't know what to do.....apparently......there was nothing to "refresh"....!!!)

 

 Nov 16, 2014
 #9
avatar+105604 
0

Thanks Chris,

 

Yes, there was a lot of 'refreshing' happening for me on these questions too.    

 

It is good to have a challenge once in a while and we both learned from it    

 Nov 17, 2014

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