Question

For what values of $a \in \mathbb{Z}$, the quadratic expression $(x + a)(x + 1991) + 1$ can be factorised as $(x + b)(x + c)$, where $b, c \in \mathbb{Z}$?

• A. $1990$
• B. $1989$
• C. $1991$
• D. $1992$

Hint:

Compare coefficient identities when matching polynomial expressions to deduce unknowns.

Answer 1

The Correct Option is B

We are given: $(x + a)(x + 1991) + 1 = x^2 + (a + 1991)x + (a \cdot 1991 + 1)$
Let this be factorised as $(x + b)(x + c) = x^2 + (b + c)x + bc$
So we equate: $a + 1991 = b + c \quad \text{and} \quad a \cdot 1991 + 1 = bc$
Let’s subtract the identity: $bc = a \cdot 1991 + 1 = (b + c - 1991) \cdot 1991 + 1$
Solve for integer values and find when both conditions hold. Try $a = 1989$:
Then expression becomes $(x + 1989)(x + 1991) + 1 = x^2 + 3980x + 1989 \cdot 1991 + 1$
This is a perfect integer factorisation. Thus, $a = 1989$ is correct.