# Thread: matching the terms in partial fractions

1. ## matching the terms in partial fractions

does anybody know where this "1" comes from?

2. ## Re: matching the terms in partial fractions

expanding the RHS yields the quadratic terms $\displaystyle 2Au^2 + Cu^2 = (2A+C)u^2$

note the coefficient of $\displaystyle u^2$ on the LHS is 1 , hence by matching coefficients $\displaystyle 2A+C = 1$

3. ## Re: matching the terms in partial fractions

u2 = (1)(u2), that is, in the polynomial u2, the coefficient of u2 is 1 (this is usually suppressed when we write polynomials).

4. ## Re: matching the terms in partial fractions

Thank you very much.
Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

5. ## Re: matching the terms in partial fractions

Originally Posted by cxz7410123
Thank you very much.
Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

LHS is $\displaystyle {\color{red}0}x^2 + 0x + 3$

you're matching coefficients, remember?