If the solutions of p(x) are -14 and 11, which function could be p?

Answer 1

If -14 and 11 are solutions to the polynomial p(x), then x = -14 and x = 11 are its roots. This means that (x + 14) and (x - 11) must be factors of the polynomial p(x). Thus, we can construct a polynomial p(x) by multiplying these factors together:

p(x) = (x + 14)(x - 11).

Expanding this we get:

p(x) = x^2 - 11x + 14x - (14 * 11)
p(x) = x^2 + 3x - 154.

So, one corresponding polynomial function could be p(x) = x^2 + 3x - 154.

Remember, any non-zero constant multiple of this polynomial is also a valid p(x). Thus, for some constant \( k \neq 0 \), the function could also be:

p(x) = k*(x^2 + 3x - 154).

This will have the same roots of -14 and 11.