2. 已知 $x = \dfrac{3}{8} + \dfrac{\sqrt{71}}{8}\mathrm{i}$ 是方程

的一个根，

(a) 直接写出该方程的另一个复数根。

已知 $x = 4$ 也是该方程的一个根，

(b) 求 $p$ 和 $q$ 的值。

原方程的系数均为实数，因此非实复根必成共轭对。把已知复根的虚部变号即可得到另一个复根。

Since the polynomial has real coefficients, its non-real roots occur in conjugate pairs. Therefore, the other complex root is

先把一对共轭复根写成两个一次因式并相乘，得到实系数二次因式。再乘第三个根 $4$ 所对应的因式 $(x-4)$，展开后与题目给出的三次多项式比较系数。

The quadratic factor corresponding to the conjugate pair is

Including the root $4$, the monic polynomial is

Multiplying by $4$ gives

Comparing coefficients with $4x^3-19x^2+px+q$,

直接使用三次方程的韦达定理。先求共轭根的和与积，再结合第三个根 $4$，由三根的积求 $q$，由两两乘积之和求 $p$。

Let the conjugate roots be $u$ and $v$. Then

After dividing the polynomial by $4$, Vieta’s formula for the product of all three roots gives

Hence

The sum of the pairwise products gives

Therefore,

利用 $x=4$ 是根，把原多项式除以 $(x-4)$。余数必须为零，而所得二次商式的两个根正是共轭复根；将商式的根积与这对共轭根的积比较，即可先求 $p$，再由余数求 $q$。

Let

When dividing by $x-4$, subtracting $4x^2(x-4)$ leaves

The next quotient term is $-3x$. Subtracting its product with $x-4$ gives

Finally, subtracting $(p-12)(x-4)$ leaves the remainder $q+4(p-12)$. Thus

Since $x=4$ is a root, the remainder is zero, so

The roots of the quadratic quotient are the conjugate pair. Their product is therefore

Hence $p=17$. Substituting this into the remainder equation gives

so

直接把已知复根代入原多项式。先算出该复根的平方与立方，再令实部与虚部分别为零，得到关于 $p,q$ 的两个实数方程。

Let

Then $s^2=71$ and $r$ is the given root. First,

Hence

Multiplying by $r$ gives

Therefore,

Since $r$ is a root,

Substituting the values above,

Equating imaginary parts gives

Since $s=\sqrt{71}\ne0$, this gives $p=17$.

Equating real parts,

Thus