Question

It is given that the polynomial $P(x) = x^3 + ax^2 + bx + c$ has three distinct positive integer roots and $P(22) = 21$.

Let $Q(x) = x^2 - 2x + 22$. It is also given that $P(Q(x))$ has no real roots then $a$ is equal to :

• A. -45
• B. -55
• C. 45
• D. 60

Answer 1

Answer: (B)

-55

Answer 2

We start with

$$P(x)=x^3+ax^2+bx+c=(x-r)(x-s)(x-t),$$

where the three distinct positive‐integer roots $r,s,t$ satisfy

$$P(22)=(22-r)(22-s)(22-t)=21.$$

Since 21 has the (only) factorization into three distinct positive integers

$$21=1\cdot 3\cdot 7,$$

one may (up to order) equate

$$22-r=1,\quad 22-s=3,\quad 22-t=7.$$

This gives

$$r=22-1=21,\quad s=22-3=19,\quad t=22-7=15.$$

(There is no other option in positive integers making the product 21 and keeping the roots distinct.)

Next, we are given

$$Q(x)=x^2-2x+22.$$

Its minimum is attained at $x=1$ (vertex) and is

$$Q(1)=1-2+22=21.$$

Now the composition

$$P(Q(x))=(Q(x)-r)(Q(x)-s)(Q(x)-t)$$

has a real solution if for any root, say $r$, the equation

$$Q(x)=r$$

has a real solution. Since

$$Q(x)=r \quad\Longleftrightarrow\quad x^2-2x+(22-r)=0,$$

its discriminant is

$$\Delta=4-4(22-r)=4r-84.$$

For there to be no real solution we need $\Delta<0$, i.e.

$$r<21.$$

Thus, if all three $r,s,t$ were strictly less than 21 then each quadratic $Q(x)=\text{root}$ would have no real solution (since $Q(x)\ge 21$ for all $x$). In our case one of the roots is $r=21$ so that (in the factor $Q(x)-21$)

$$Q(x)=21 \quad\Longrightarrow\quad (x-1)^2=0,$$

giving a real (repeated) solution.

Since the choices are predetermined and the only possible choice (forced by $P(22)=21$) leads to

$$r=21,\quad s=19,\quad t=15,$$

the value of

$$a=-(r+s+t)=-\big(21+19+15\big)=-55.$$

Thus, among the options provided, the answer is option (B) – even though note that strictly speaking the condition “$P(Q(x))$ has no real roots” is violated by $x=1$. In a multiple–choice context the only candidate is $-55$.