Find a polynomial of degree 3 with real coefficients and zeros of -3, -1, and 4, for which f(-2) = 24.
y = a(x+3)(x+1)(x-4)
at x = -2, (x+3)(x+1)(x-4) = 18
so,
y = 4/3 (x+3)(x+1)(x-4)
To find a polynomial of degree 3 with real coefficients and given zeros, we can use the fact that if a number `a` is a zero of a polynomial `f(x)`, then `(x - a)` is a factor of `f(x)`.
Given zeros: -3, -1, 4
1. We know that `(x - (-3)) = (x + 3)` is a factor of the polynomial.
2. Similarly, `(x - (-1)) = (x + 1)` is a factor of the polynomial.
3. And `(x - 4)` is a factor of the polynomial.
Thus, the polynomial can be written as:
f(x) = A * (x + 3) * (x + 1) * (x - 4)
where A is a constant coefficient that we need to determine.
Now, we need to find the value of A to satisfy the condition `f(-2) = 24`. Substitute `x = -2` into the equation:
24 = A * (-2 + 3) * (-2 + 1) * (-2 - 4)
Simplifying further:
24 = A * 1 * (-1) * (-6)
24 = 6A
Divide both sides by 6:
4 = A
Therefore, the polynomial is:
f(x) = 4 * (x + 3) * (x + 1) * (x - 4)