Question

Find a third–degree polynomial function such that f(0) = –24 and whose zeros are –1, 2, and –3.

Answer 1

This requires generating a systems of equations. List out three polynomial functions of third-degree: f(x) = ax^3 + bx^2 + cx + d To find the four unknown coefficients of your third-degree polynomial, you need four equations. f(0) = -24 is one equation.

f(0) = f(x=0) = a(0)^3 + b(0)^2 + c(0) + d = d = -24

Therefore: d = -24.

Hint to find the other coefficients: The definition of a "zero" is f(x) = 0

Answer 2

Thank you:)

Answer 3

If -1, 2, and 3 are your zeros, then x-(-1), x-2 and x-3 are factors of your third degree polynomial. Also, f(0)=-24, so
F(X)=a(x+1)(x-2)(x-3) is what your polynomial looks like. Then substitute x=0,
and we have f(0)=-24=a(0+1)(0-2)(0-3), therefore,
-24=a(6).
Solving for a, we get a=-4. This is your leading coefficient of your polynomial. Your equation is then,
f(x)=-4(x+1)(x-2)(x-3). Multiply the factors out, and you will get your your polynomial of degree three with the required zeros.

Answer 4

Thank you:) both @Aeronautix and @calmat01!!! You both explained every well:)

Answer 5

Np.

Answer 6

You are most welcome.

Answer 7

calmat's answer is simpler, use his approach. :) But also know the concept of systems of equations when the function is not a polynomial!

Answer 8

Will do!:)