Đáp án`+`Giải thích các bước giải:
`a)` \begin{array}{c|c} \quad \quad \quad x^4+ax^2+b&x^2-x+1\\\hline x^4-x^3+x^2&x^2+x+a\\\text{_____________________}\\\quad \quad x^3+(a-1)x^2+b\\x^3-x^2+x\\\text{_____________________}\\ax^2-x+b \\ax^2-ax+a\\\text{_____________________}\\(a-1)x+b-a\\\end{array}
Để `x^4+ax^2+b` chia hết cho `x^2-x+1`
thì `(a-1)x+b-a=0`
`=>{(a-1=0),(b-a=0):}<=>{(a=1),(b-1=0):}<=>{(a=1),(b=1):}`
Vậy `a=b=1` để `x^4+ax^2+b` chia hết cho `x^2-x+1`
`b)` Ta có:
`x^2+3x-10`
`=x^2+5x-2x-10`
`=x(x+5)-2(x+5)`
`=(x+5)(x-2)`
`=>` $\left[\begin{matrix} x+5=0\\ x-2=0\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=-5\\ x=2\end{matrix}\right.$
Đặt `f(x)=ax^3+bx^2+5x-50`
Để `f(x)=ax^3+bx^2+5x-50` chia hết cho `x^2+3x-10`
thì `f(-5)=f(2)=0`
`<=>{(f(-5)=-125a+25b-25-50=0),(f(2)=8a+4b+10-50=0):}<=>{(-125a+25b-75=0),(8a+4b-40=0):}<=>{(-125a+25b=75),(8a+4b=40):}<=>{(-25(5a+b)=75),(4(a+2b)=40):}<=>{(5a+b=-3),(a+2b=10):}<=>{(a=\frac{-13}{3}),(b=\frac{56}{3}):}`
Vậy `a=-13/3;b=56/3`
