Question

Find a polynomial function of degree 3 such that f(10)=17 and the zeros of f are 0, 5 and 8

Answer 1

Step-by-step explanation:

Since f(0) = f(5) = f(8) = 0, we have f(x) = Ax(x - 5)(x - 8), where A is a real constant.

We know that f(10) = 17.

=> A(10)(10 - 5)(10 - 8) = 17

=> A(10)(5)(2) = 17

=> 100A = 17, A = 0.17.

Hence the answer is f(x) = 0.17x(x - 5)(x - 8).