Thread: matching the terms in partial fractions

does anybody know where this "1" comes from?

expanding the RHS yields the quadratic terms

note the coefficient of on the LHS is 1 , hence by matching coefficients

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

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

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

you're matching coefficients, remember?