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(x^2+mx+n)(x^2-3x+2) =x^4-3x^3+2x^2+mx^3-3mx^2+2mx+nx^2-3nx+2n =x^4+(m-3)x^3+(2-3m+n)x^2+(2m-3n)x+2n 因為不含x^2和x項 所以 2-3m+n=0 2m-3n=0 解之得, m=6/7, n=4/7
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