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.
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
$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\
f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$
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
$$\\\frac{2x-13}{(2x+1)(x-3)}\\\\
=\frac{A}{2x+1}-\frac{B}{x-3}\\\\
=\frac{A(x-3)-B(2x+1)}{(2x+1)(x-3)}\\\\
=\frac{Ax-3A-2Bx-B}{(2x+1)(x-3)}\\\\
=\frac{(A-2B)x-(3A+B)}{(2x+1)(x-3)}\\\\
so\\\\
A-2B=2\qquad and \qquad 3A+B=13\\
A=2+2B\\
3(2+2B)+B=13\\
6+7B=13\\
7B=7\\
B=1\\
A=2+2*1\\
A=4\\
so\\\\
\frac{2x-13}{(2x+1)(x-3)}=\frac{4}{2x+1}-\frac{1}{x-3}$$
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. :)
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
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.
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
$$\\f(x)=\frac{x^2-10}{(x-2)(x+1)}\\\\
f(x)=1\;+\;\frac{x-9}{(x-2)(x+1)}\\\\
f(x)=1\;-\;\frac{7/3}{x-2}+\frac{10/3}{x+1}\\\\$$