Đặt `f(x)=ax^3+bx^2+5x-50`
Có `x^2+3x-10=(x+5)(x-2)`
`f(x) vdots (x^2+3x-10)`
`<=>f(x) vdots (x+5)(x-2)`
`<=> f(-5)=f(2)=0`
`<=>-125a+25b-25-50=8a+4b+10-50=0`
`<=>-125a+25b=75 ; 8a+4b=40`
`<=>-5a+b=3;2a+b=10`
`<=>9a=7;2a+b=10`
`<=>a=7/9;14/9+b=10`
`<=>a=7/9;b=90/9-14/9=86/9`