Write the factored form for the polynomial function that has zeros of 2, -3, and 5; and f(3)=6.

Answer 1

P(x) = a(x - 2)(x + 3)(x - 5)

P(3) = 6

6 = a(3 - 2)(3 + 3)(3 - 5)

Can you solve for a and answer?

Answer 2

If roots to a polynomial function f(x) are given, in this case, 2,-3 and 5, that means we can write the polynomial in the form f(x) = ( x - root 1)(x - root 2)(x - root3) or f(x) - (x-2)(x - (-3))(x - 5) or f(x) = (x-2)(x+3)(x-5). There are an infinite polynomial functions with these 3 roots, but the coefficient "a" is what makes the roots be what they are AND, uniquely, f(3) = 6. Remember f(3) = 6 means that the point ( 3,6) lies on the graph. So "a" will go in front of the factors. Let's expand the above. We have f(x) = a(x-2)(x+3)(x-5). We have f(x) = a(x^3 - 4x^2 -11x + 30). [ Remember that f(x) = 6 means that when x = 3, y = 6. ] So....let's sub in that (x,y) value and solve for "a". a(3^3 -4(3)^2 -11(3) + 30) = 6 a(27-4(9) -33 + 30) =6 a(27-36-33+30) =6 a(-12) = 6 -12a = 6 a = -.5 or - 1/2 So our polynomial is f(x) = -(1/2)( x^3 -4x^2 - 11x+30) or f(x) = (1/2)x^3 -2x^2 -(11/2)x +15 Enter this equation in the graphing calculator and look at the tables. You will see the three roots (2,0), (-3,0) (5,0) and f(3) =6 expressed as (3,6)