Create a cubic function

Question

Create a cubic function that has f(-2) f(3) f(1) = 0 and f(2) = 12. Keep answer factored.

Answer 1

The simplest polynomial satisfying f(c) = 0 for some c value is the function f(x) = x - c. Since zero times anything is zero, we can take the product of simple polynomials like this to get a new polynomial that still equals zero at each given point. Another way to think of this: if f(c) = 0 at some value c, then x - c will be a factor of f(x).

So, a possible function that has zeroes at -2, 3, and 1 is

f(x) = (x+2)(x-3)(x-1).

If we plug 2 into this function, we get f(2) = (4)(-1)(1) = -4, so this can't be our final answer. However, we can use the fact that zero times anything is zero again. If f(c) = 0 at some value, and we multiply f(x) by some constant M, then Mf(c) = M·0 = 0 (i.e., the function still has a zero at c). That means if M is some real number, then any polynomial of the form

f(x) = M(x+2)(x-3)(x-1)

has zeroes at -2, 3, and 1. From here, we note, f(2) = M(4)(-1)(1) = -4M. So, to satisfy the condition that

f(2) = -4M = 12,

we must have that M = -3. Hence,

f(x) = -3(x+2)(x-3)(x-1)

is a cubic polynomial such that f(-2), f(3), f(1) = 0 and f(2) = 12.

Answer 2

If f(c) = 0, then x - c is a factor of f(x).

Since f(-2) = f(3) = f(1) = 0, x+2, x-3, and x-1 are factors of f(x).

f(x) = A(x+2)(x-3)(x-1)

Since f(2) = 12, A(4)(-1)(1) = 12. So, A = -3.

Therefore, f(x) = -3(x+2)(x-3)(x-1)

Answer 3

Hello Bobella,

Assuming you meant f(-2) = 0, f(3) = 0 ,f(1) = 0 and f(2) = 12:

f(x) = A (x+2)(x-3)(x-1) where A is the leading coefficient.

To find A - you'd want to use f(2) = 12: plug 2 wherever x is, replace f(x) by 12 and solve for A:

12 = A (2+2)(2-3)(2-1) --> 12 = A (4)(-1)(1) ---> 12 = -4A so A = -3

Your final answer for f(x) can thus be written as: f(x) = -3(x+2)(x-3)(x-1)