does anybody know where this "1" comes from?
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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 $\displaystyle 2A+C = 1$
u^{2} = (1)(u^{2}), that is, in the polynomial u^{2}, the coefficient of u^{2} 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 $\displaystyle {\color{red}0}x^2 + 0x + 3$ you're matching coefficients, remember?
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