Help w/ decomposing (17n-23) / (4n^2+23n-6)

[Jen123]

Use partial functions to decompose the expression: (17n-23) / (4n^2+23n-6)

first I factored the denominator: (4n-1) + (1n+6)

I expresses the factored form as the sum of two fractions using A and B as numerators and the factors as denominators: A /(4n-1) + B /(1n+6)

I eliminated the denominator by multiplying by the LCD (4n-1)(1n+6):
A(1n+6) + B(4n-1)

Then I have some problems because in my book it says eliminate B by letting (4n-1) equal zero. But I don't know what to multiply 4 by to let the factor equal zero.

When I eliminated A I made n=-6 and got B=-5.43 or -125/23 but I'm not sure if this is right.

Some help would be useful

 

[pka]

\(\displaystyle \L \frac{5}{{n + 6}} - \frac{3}{{4n - 1}}\)

 

[Steem]

Then I have some problems because in my book it says eliminate B by letting (4n-1) equal zero. But I don't know what to multiply 4 by to let the factor equal zero.

Wussup Jen123,

You can eliminate B by letting n = 1/4. Solve for A then do the same to solve for B (eliminate A).

 

[Jen123]

thanx 4 the help