\(\left(x^2-3x+2\right)\left(x^2+cx+d\right)\)
\(=x^4+c\cdot x^3+dx^2-3x^3-3c\cdot x^2-3x\cdot d+2x^2+2cx+2d\)
\(=x^4+x^3\left(c-3\right)+x^2\left(d-3c+2\right)+x\left(-3d+2c\right)+2d\)
Theo đề, ta có: c-3=0; d-3c+2=a; -3d+2c=0; 2d=b
=>c=3; d+2-9=a; 3d=2c=6; 2d=b
=>c=3; d=2; b=4; a=2+2-9=-5