Bài 3:
`a) x^3-5x^2+11x-10`
`=x^3-2x^2-3x^2+6x+5x-10`
`=x^2(x-2)-3x(x-2)+5(x-2)`
`=(x-2)(x^2-3x+5)`
`-> f(x):g(x)=x^2-3x+5`
`b) 3x^3-7x^2+4x-4`
`=3x^3-6x^2-x^2+2x+2x-4`
`=3x^2(x-2)-x(x-2)+2(x-2)`
`=(x-2)(3x^2-x+2)`
`-> f(x):g(x)=3x^2-x+2`
Bài 4:
`a) x^3+ax^2+2x+b`
`=x^3+x^2+x+(a-1)x^2+(a-1)x+(a-1)-(a-1)x+2x-(a-1)+b`
`=x(x^2+x+1)+(a-1)(x^2+x+1)+(2-a+1)x+b-a+1`
`=(x^2+x+1)(x+a-1)+(3-a)x+(b-a+1)`
Để `x^3+ax^2+2x+b \vdots x^2+x+1`
`=> (3-a)x+(b-a+1) \vdots x^2+x+1`
`<=> {(3-a=0),(b-a=-1):}<=>{(a=3),(b=-1+3):}<=>{(a=3),(b=2):}`
Vậy `(a;b)=(3;2)`
`b) x^4-9x^3+21x^2+x+a`
`=x^4-x^3-2x^2-8x^3+8x^2+16x+15x^2-15x-30+a+30`
`=x^2(x^2-x-2)-8x(x^2-x-2)+15(x^2-x-2)+a+30`
`=(x^2-x-2)(x^2-8x+15)+a+30`
Để `x^4-9x^3+21x^2+x+a \vdots x^2-x-2`
`=> a+30 \vdots x^2-x-2`
`<=> a+30=0`
`<=> a=-30`
Vậy `a=-30`
`c) x^4-3x^3+3x^2+ax+b`
`=x^4-3x^3+4x^2-x^2+3x-4+(a-3)x+b+4`
`=x^2(x^2-3x+4)-(x^2-3x+4)+(a-3)x+b+4`
`=(x^2-3x+4)(x^2-1)+(a-3)x+b+4`
Để `x^4-3x^3+3x^2+ax+b \vdots x^2-3x+4`
`=> (a-3)x+b+4 \vdots x^2-3x+4`
`<=> {(a-3=0),(b+4=0):}<=>{(a=3),(b=-4):}`
Vậy `(a;b)=(3;-4)`
