• Nov 26th 2007, 01:24 PM
soly_sol
Write the partial fraction decomposition of the rational expression.

3. (9x^2-x-14)/(x^3-x)

I got: (14)/(x) +(-2)/(x+1) + (-3)/(x-1)

Write the partial fraction decomposition of the rational expression.

11. (6x+5)/((x-9)^2)

I got: (6)/(x-9) + (59)/(x-9)^2

13. (5x^2-5x+7)/((x-1)^3)
I got: (5)/(x-1) - (5)/(x-1)^2 + (7)/(x-1)^2

15. (x+4)/(x^3-2x^2+x)
I got: (4)/(x) + (5)/(x-1) + (-4)/(x-1)^2

Write the partial fraction decomposition of the rational expression.

17. (12x+3)/ (x^3-1)
I got: (-5)/(x-1) + (5x+2)/(x^2+x+1)

(12x^2 -12x+6)/(x-3)(x^2+4)
I got: (6)/(x-3) + (6x+6)/(x^2+4)

Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants.

26. (7x-3)/((x^2+x-7)^2)

I got: (ax+b)/(x^2+x-7) + (cx+d)/((x^2+x+7)^2)

29. (2x+1)/(x-6)((x^2+x-4)^2)

I got: (a)/(x-6) + (bx+c)/(x^2+x-4)

Write the partial fraction decomposition of the rational expression.

31. (x^2+4x-1)/((x^2+3)^2)
I got: (1)/(x^2-3) + (4x-4)/(x^2+3)^2

33. (4x^3+ 4x^2)/((x^2+5)^2))
I got: (4x+4)/(x^2+5) + (-20x-20)/((x^2+5)^2))
• Nov 26th 2007, 05:00 PM
Soroban
Hello, soly_sol!

Quote:

Write the partial fraction decomposition of the rational expression.

$\displaystyle 3)\;\;\frac{9x^2-x-14}{x^3-x}$

I got: .$\displaystyle \frac{14}{x} - \frac{2}{x+1} -\frac{3}{x-1}$ . . . . Right!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle 11)\;\;\frac{6x+5}{(x-9)^2}$

I got: .$\displaystyle \frac{6}{x-9} + \frac{59}{(x-9)^2}$ . . . . Yes!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle 13)\;\;\frac{5x^2-5x+7}{(x-1)^3}$

I got: .$\displaystyle \frac{5}{x-1} - \frac{5}{(x-1)^2} + \frac{7}{(x-1)^2}$ . . . . no

It should be: .$\displaystyle \frac{5}{x-1} + \frac{5}{(x-1)^2} + \frac{7}{(x-1)^3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle 15)\;\;\frac{x+4}{x^3-2x^2+x}$

I got: .$\displaystyle \frac{4}{x} + \frac{5}{x-1} - \frac{4}{(x-1)^2}$ . . . . no

It should be: .$\displaystyle {\color{blue}\frac{4}{x} - \frac{4}{x-1} + \frac{5}{(x-1)^2}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle 17)\;\;\frac{12x+3}{x^3-1}$

I got: .$\displaystyle \frac{-5}{x-1} + \frac{5x+2}{x^2+x+1}$ . . . . no

It should be: .$\displaystyle {\color{blue}\frac{5}{x-1} + \frac{-5x+2}{x^2+x+1}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

. . You should get the original fraction.

In #13, they gave us: .$\displaystyle \frac{5x^2 - 5x + 7}{(x-1)^3}$

You came up with: .$\displaystyle \frac{5}{x-1} - \frac{5}{(x-1)^2} + \frac{7}{(x-1)^3}$

Let's check it . . . get a common denominator and combine.

$\displaystyle \frac{5}{x-1}\cdot{\color{green}\frac{(x-1)^2}{(x-1)^2}} \:- \:\frac{5}{(x-1)^2}\cdot{\color{green}\frac{x-1}{x-1}} \:+ \:\frac{7}{(x-1)^3}$

. . . $\displaystyle = \;\frac{5(x-1)^2}{(x-1)^3} \:- \:\frac{5(x-1)}{(x-1)^3} \:+\: \frac{7}{(x-1)^3} \;\;=\;\;\frac{5(x-1)^2 - 5(x-1) + 7}{(x-1)^3}$

. . . $\displaystyle = \;\frac{5x^2 - 10x + 5 - 5x + 5 + 7}{(x-1)^3} \;\;=\;\;\frac{5x^2 - {\color{red}15}x + {\color{red}17}}{(x-1)^3}$ ??

This is not the original fraction . . . The solution is wrong.

See how it works?