$f(x)=(x-2)(2x^2+5x+k)+21$，其中 $k$ 是常数。

(a) 写出 $f(x)$ 除以 $(x-2)$ 的余数。（1分）

已知 $(2x-1)$ 是 $f(x)$ 的一个因式。

(b) 证明 $k=11$。（2分）

(c) 因此：

(i) 将 $f(x)$ 完全因式分解；

(ii) 求方程 $f(x)=0$ 的实数解的个数，并说明理由。（共5分）

当 $x=2$ 时，$(x-2)$ 那一整项为 $0$，所以余数就是 $f(2)$。

By the remainder theorem, the remainder is $f(2)$.

Therefore, the remainder is

$(2x-1)$ 是因式，所以 $x=\frac12$ 是根。代入 $f(x)=0$ 解 $k$。

Since $(2x-1)$ is a factor,

So

Thus

This is the required result.

先将含 $k$ 的三次式除以已知因式 $2x-1$。余数必须为零，据此求 $k$，所得商式还可承接第(c)问。

First expand:

Dividing by $2x-1$, the first quotient term is $x^2$:

The next quotient term is $x$:

The final quotient term is $\frac{k-9}{2}$:

Since $2x-1$ is a factor, the remainder is zero:

The quotient is $x^2+x+\frac{k-9}{2}$, which becomes $x^2+x+1$ when $k=11$. Thus this division also gives the factorisation required in part (c)(i).

先把 $k=11$ 代入并展开成三次式，再利用已知因式 $(2x-1)$ 分解。

With $k=11$,

Divide by $2x-1$. The quotient terms are successively $x^2$, $x$ and $1$:

The remainder is zero, so

Therefore,

待定系数法（inspection）。代入 $k=11$ 展开得到三次式 $f(x) = 2x^3+x^2+x-1$。因为已给因式 $(2x-1)$，可设商式为二次式 $(ax^2+bx+c)$。通过两端的最高次项与常数项可以直接口算出 $a$ 和 $c$，再对比中间项系数得到 $b$，免去多项式长除法。

Substitute $k=11$ and expand the cubic:

Since $(2x-1)$ is a factor of $f(x)$, we can assume the factorised form is:

Compare the coefficients:

• For the $x^3$ term:

• For the constant term:

• For the $x^2$ term (expanding $(2x-1)(x^2+bx+1)$):

Therefore, the quadratic factor is $(x^2+x+1)$, and the fully factorised form of $f(x)$ is:

线性因式给一个实根。二次因式是否有实根，要看判别式。

From part (c)(i),

The factor $2x-1=0$ gives

For $x^2+x+1=0$, the discriminant is

Since $-3<0$, the quadratic has no real roots.

Therefore, $f(x)=0$ has

real solution.

同一二次因式也可以用配方判断：

for every real $x$. Thus it cannot contribute a real root, and $x=\frac12$ is the only real solution.