\(x^4-7x^3+4x^2+ax+b-1⋮x^2-7x+3\)
\(\Rightarrow x^4-7x^3+4x^2+ax+b-1=\left(x^2-7x+3\right)\left(x^2+cx+d\right)\left(1\right)\)
\(\left(x^2-7x+3\right)\left(x^2+cx+d\right)=x^4+cx^3+dx^2−7x^3−7cx^2−7dx+3x^2+3cx+3d\)
\(=x^4+(c−7)x^3+(d−7c+3)x^2+(3c−7d)x+3d\)
\(\left(1\right)\Rightarrow x^4-7x^3+4x^2+ax+b-1=x^4+(c−7)x^3+(d−7c+3)x^2+(3c−7d)x+3d\)
Đồng nhất thức các hệ số ta được :
\(\left\{{}\begin{matrix}c-7=-7\left(2\right)\\d-7c+3=4\left(3\right)\\3c-7d=a\left(4\right)\\3d=b-1\left(5\right)\end{matrix}\right.\)
\(\left(2\right)\Rightarrow c=0\)
\(\left(3\right)\Rightarrow d=1\)
\(\left(4\right)\Rightarrow a=-7\)
\(\left(5\right)\Rightarrow b=4\)
Vậy \(\left(a;b\right)=\left(-7;4\right)\) thỏa mãn đề bài
