Q. Which Function Has Real Zeros at \( x = 3 \) and \( x = 7 \)?

Problem

Which function has real zeros at \( x = 3 \) and \( x = 7 \)?

Step-by-step solution

1. Interpret the meaning of a zero (root):

   If a function \( f(x) \) has a real zero at \( x = a \), then \( x – a \) is a factor of \( f(x) \). Therefore, zeros at \( x = 3 \) and \( x = 7 \) mean the factors \( x – 3 \) and \( x – 7 \) must appear in the function.

2. Write a general polynomial with those factors:

   The most general real polynomial with those zeros (and with those zeros having multiplicity 1) is

   \( f(x) = c\,(x – 3)(x – 7) \)

   where \( c \) is any nonzero real constant. Choosing \( c = 1 \) gives the simplest such function.

3. Give the simplest explicit function:

   Choose \( c = 1 \). Then

   \( f(x) = (x – 3)(x – 7) \)

   Expand the product:

   \( f(x) = x^2 – 7x – 3x + 21 \)

   \( f(x) = x^2 – 10x + 21 \)

4. Verify the zeros:

   Evaluate at \( x = 3 \):

   \( f(3) = (3 – 3)(3 – 7) = 0 \)

   Evaluate at \( x = 7 \):

   \( f(7) = (7 – 3)(7 – 7) = 0 \)

   Both evaluations equal zero, so the function has the required real zeros.

Answer

A simplest function with real zeros at \( x = 3 \) and \( x = 7 \) is

\( f(x) = (x – 3)(x – 7) = x^2 – 10x + 21 \).

More generally, any nonzero constant multiple \( f(x) = c\,(x – 3)(x – 7) \) with \( c \in \mathbb{R}\setminus\{0\} \) will have those zeros.