多项式 $2x^4+ax^3+bx-1$ 记为 $\mathrm{p}(x)$，其中 $a$ 和 $b$ 为常数。$\mathrm{p}(x)$ 除以 $x^2-x+1$ 所得的余式为 $3x+2$。

求 $a$ 和 $b$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

进行含参多项式除法，把余式写成一次式；再逐项与已知余式比较。

Dividing by $x^2-x+1$, the first quotient term is $2x^2$. Subtraction leaves

The next quotient term is $(a+2)x$. Subtraction leaves

The final quotient term is $a$. Subtraction leaves

Thus

Comparing the remainder with $3x+2$ gives

Hence $\boxed{a=-3,\qquad b=5}$.

也可从“被除式等于除式乘商加余式”出发，设二次商并比较系数，避免含参长除法。

The quotient has leading term $2x^2$. Write

Expanding the right side,

Comparing constant terms gives $B+2=-1$, so $B=-3$. The coefficient of $x^2$ gives $B-A+2=0$, so $A=-1$.

Therefore

Hence $\boxed{a=-3,\qquad b=5}$.

取除式的一个非实根，商乘除式的部分便为零。利用该根满足的二次方程，先化简三次、四次幂，再比较实部与虚部。

Let $w=(1+\sqrt3\,\mathrm{i})/2$, a root of $w^2-w+1=0$. Then

Since the remainder is $3x+2$, the polynomial identity gives $\mathrm p(w)=3w+2$. Hence

The right side is real. Since $\operatorname{Im}w=\sqrt3/2\ne0$ and $a,b$ are real, comparing imaginary parts gives $b-5=0$. The real parts then give $a+3=0$. Thus $\boxed{a=-3,\quad b=5}$.