Đặt \(f\left(x\right)=ax^3+bx^2-11x+30\)
Ta có : \(x^2-3x-10=\left(x+2\right)\left(x-5\right)\)
+) \(f\left(x\right)⋮x+2\)
\(\Leftrightarrow f\left(-2\right)=0\)
\(\Leftrightarrow-8a+4b-11.\left(-2\right)+30=0\)
\(\Leftrightarrow-8a+4b+22+30=0\)
\(\Leftrightarrow-8a+4b+52=0\)
\(\Leftrightarrow-2a+b+13=0\)( * )
+) \(f\left(x\right)⋮x-5\)
\(\Leftrightarrow f\left(5\right)=0\)
\(\Leftrightarrow125a+25b-11.5+30=0\)
\(\Leftrightarrow125a+25b-25=0\)
\(\Leftrightarrow5a+b-1=0\)
\(\Leftrightarrow-2a+7a+b+13-14=0\)
\(\Leftrightarrow\left(-2a+b+13\right)+\left(7a-14\right)=0\)( ** )
Từ ( * ) ; ( ** )
\(\Rightarrow7a-14=0\)
\(\Rightarrow7a=14\)
\(\Rightarrow a=2\)
\(\Rightarrow b=-9\)
Vậy với \(a=2;b=-9\) thì \(ax^3+bx^2-11x+30⋮x^2-3x-10\)
