Find a polynomial function of least degree with real coefficients satisfying the given properties.
zeros -3, 0, and 4
f(1) =10
if zeros are -3, 0, 4, it could be
f(x) = x(x+3)(x-4)
but (1,10) lies on it , so we have to stretch it to pass through that point without wrecking the zeros
f(x) = ax(x+3)(x-4)
10 = a(4)(-3)
-12a = 10
a = -5/6
f(x) = -(5/6)x(x+3)(x-4)
Thank you.
To find a polynomial function with the given zeros and value, we know that the factors of the polynomial will be (x - a)(x - b)(x - c), corresponding to the zeros -3, 0, and 4.
So, the polynomial will be of the form f(x) = (x - (-3))(x - 0)(x - 4).
To determine the degree of the polynomial, we count the number of factors, which is 3 in this case. Therefore, the degree of the polynomial will be 3.
Expanding the expression, we get f(x) = (x + 3)(x)(x - 4).
Now, we can simplify and find the polynomial function:
f(x) = (x + 3)(x)(x - 4)
= (x^2 + 3x)(x - 4)
= x^3 - 4x^2 + 3x^2 - 12x
= x^3 - x^2 - 12x
So, the polynomial function of least degree with real coefficients that satisfies the given properties is f(x) = x^3 - x^2 - 12x.