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:

When comparing two quadratic expressions, equate coefficients of corresponding powers of \(x\). Completing factor-type rearrangements often helps simplify integer factorization problems.

Answer 1

The Correct Option is B

Step 1: Expand the given quadratic expression.
We have \[ (x+a)(x+1991)+1 \] Expanding, \[ =x^2+(a+1991)x+1991a+1 \] We are given that it can be factorised as \[ (x+b)(x+c) \] Expanding the factorised form, \[ =x^2+(b+c)x+bc \] Comparing coefficients, we get \[ b+c=a+1991 \] and \[ bc=1991a+1 \]

Step 2: Express \(a\) in terms of \(b\) and \(c\).
From \[ b+c=a+1991, \] we get \[ a=b+c-1991 \] Substituting into \[ bc=1991a+1, \] we obtain \[ bc=1991(b+c-1991)+1 \] \[ bc=1991b+1991c-1991^2+1 \] Rearranging, \[ bc-1991b-1991c+1991^2=1 \] Adding and subtracting \(1\), \[ (b-1991)(c-1991)=1 \]

Step 3: Use integer factorization.
Since \(b,c\in\mathbb{Z}\), \[ (b-1991)(c-1991)=1 \] The integer factors of \(1\) are: \[ 1\times1 \quad \text{or} \quad (-1)\times(-1) \] Case 1: \[ b-1991=1,\qquad c-1991=1 \] Thus, \[ b=c=1992 \] Then, \[ a=b+c-1991 \] \[ a=1992+1992-1991 \] \[ a=1993 \] This is not among the options.
Case 2: \[ b-1991=-1,\qquad c-1991=-1 \] Thus, \[ b=c=1990 \] Hence, \[ a=1990+1990-1991 \] \[ a=1989 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{1989} \] which corresponds to option (2).