Polynomial Functions

Question

write a factored form for a polynomial function of degree three that has zeros of 2, -3, and 5.

Answer 1

Since zeroes or "x-intercepts" are where the y = 0 on the graph of the function, the factored form of the polynomial must be written in the form

f(x) = (x - a)(x - b)(x - c). . .(x - _degree)

Substituting in the zeroes of 2, -3, and 5, we obtain

f(x) = (x - 2)(x + 3)(x - 5)

Our solution has a degree of 3 because the degree is the number of the highest exponent in the function, and when expanding the factored form of the polynomial, we get

f(x) = x³ - 4x² - 11x + 30

Answer 2

(x-2)(x+3)(x-5) = 0

expand

(x^2 + x -6)(x-5) = 0

x^3 -5x^2 -11x + 30 = 0

just take each zero and change it's sign, then add each to x to get 3 factors of the polynomial