decomp into partial fractions: 14x+84/((x^2)+11x+18)

[cjswonderfulgirl]

hey-i'm in college algebra, and i have a set of about six problems that i just cannot figure out. for example, i have to break down

14x+84/((x^2)+11x+18)
and i've gotten to the following steps:
14x+84/(x+9)(x+2)=A/x+9 + B/x+2
14x+84=A(x+2)+B(x+9)

substitute x=-2

14(-2)+84=A(-2+2)+B(-2+9)
-28+84=7B
B=8

substitute x=-9
14(-9)+84=A(-9+2)+B(-9+9)
126=-7A
A=-18

so, thus, 14x+84/(x+9)(x+2)=-18/x+9 +8/x+2

and i don't understand exactly what i did wrong. all my homework is due at 11:59 on Saturday nights, so i'll understand if i don't get an explanation by then, but i would really like to know what i'm doing wrong, because i thought i had it right, and i even tried to solve it changing it to A/x+2 +B/x+9 and i got the same answer. i just don't know where i went wrong, but thanks in advance for your time

 

[galactus]

Re: decomposition into partial fractions

B=8 is correct. You just need A. It is not -18.

\(\displaystyle \frac{A}{x\, +\, 9}\, +\, \frac{B}{x\, +\, 2}\, =\, \frac{14x+84}{x^2\,+\,11x\,+\,18}\)

\(\displaystyle Ax\, +\, 2A\, +\, Bx\, +\, 9B\, =\, 14x\, +\, 84\)

Like coefficients:

A + B = 14

2A + 9B =8 4

A + 8 = 14, so A = 6

 

[Deleted member 4993]

Re: decomposition into partial fractions

There is another way - which I find to be quicker

\(\displaystyle \frac{A}{x+9}+\frac{B}{x+2}=\frac{14x+84}{(x+9)(x+2)}\)

Multiply both sides by (x+9)

\(\displaystyle A+\frac{B(x+9)}{x+2}=\frac{14x+84}{x+2}\)

Now take the limit of both sides for x -> (-9)

\(\displaystyle A = \frac{14(-9)+84}{(-9)+2} = 6\)

Similarly 'B' can be found by multiplying both sides with (x+2) and taking the limit to x -> (-2)

 

[soroban]

Re: decomposition into partial fractions

Hello, cjswonderfulgirl!

One of those silly arithmetic errors . . .

Partial Fractions: .\(\displaystyle \frac{14x+84}{x^2+11x+18}\)

and i've gotten to the following steps: .\(\displaystyle \frac{14x+84}{(x+9)(x+2)}\:=\:\frac{A}{x+9} + \frac{B}{x+2}\)

. . . \(\displaystyle 14x+84\:=\:A(x+2)+B(x+9)\)

\(\displaystyle \text{Substitute }x=-2:\;\;14(-2)+84\;=\;A\cdot0 +B\cdot7\quad\Rightarrow\quad 56 = 7B\quad\Rightarrow\quad B = 8\)

\(\displaystyle \text{Substitute }x=-9:\;\;14(-9)+84\:=\:A\cdot0 + B\cdot0\quad\Rightarrow\quad \underbrace{126}_{\text{Here!}}\:=\:-7A\)

You should have had: .\(\displaystyle -42 \:=\:-7A\quad\Rightarrow\quad A = 6\)

 

[galactus]

Re: decomposition into partial fractions

Yes, indeedy. With PFD's (and most other things), meticulously look back. It's mostly a small arithmetic error. That's all it takes to throw a wrench in the gears of your plan. The Butterfly Effect of the math world.

 

[cjswonderfulgirl]

Re: decomposition into partial fractions

i understand these ones now (without squares and all that thrown around the numerator...) i have one more question, and i swear it's my last, on the board right above this. i just want to know what i did wrong there you guys are so smart! I hope someday i can help people out like you do thanks so much