Multiple choice

Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which a and b are integers. If for a given integer $n$, $f(n) f(n + 1) = f(m)$ for some integer $m$, then the value of m is

  1. $n(a + b) + ab$
  2. $n^2 + an + b$
  3. $n(n + 1) + an + b$
  4. $n^2 + n + a + b$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

f(n) = n^2 + an + b. f(n+1) = (n+1)^2 + a(n+1) + b = n^2 + 2n + 1 + an + a + b. f(n)f(n+1) = (n^2 + an + b)(n^2 + 2n + 1 + an + a + b). This looks like it should match f(m) = m^2 + am + b. By testing values or algebraic expansion, m = n(n+1) + an + b.

AI explanation

Expand the given equation f(n)f(n+1)=f(m) by substituting the polynomial x^2+ax+b. This gives (n^2+an+b)((n+1)^2+a(n+1)+b)=m^2+am+b. Observe that n^2+an+b can be written as n(n+a)+b, and the second factor can be grouped as (n+1)(n+1+a)+b. The product simplifies algebraically when we rewrite the terms by adding and subtracting 1 from the first factor, forming (n^2+n+an+b) which equals f(n)+n. This identity shows that the integer m is n(n+1)+an+b.

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