how do i construct a polynomial function with zeros, containing points, and a degree?

Question

zeros: -3 (multiplicity 2), 2(multiplicity 1), 4(multiplicity 1)

degree: 4

contains the point (1,96)

Answer 1

If "a" is a zero, x - a is a factor.

f(x) = (x + 3)(x + 3)(x - 2)(x - 4)

f(1) = (1 + 3)(1 + 3)(1 - 2)(1 - 4)

f(1) = (4)(4)(-1)(-3)

f(1) = 48

Yet the condition is f(1) = 96

 

f(x)= 2(x + 3)(x +3)(x - 2)(x - 4)

 

Answer 2

If "x" is a zero, then x-a is a factor of the polynomial equation

 

Multiplicity tells you how many times to use it as a factor

 

f(x) = a (x - -3)(x- -3)(x-2)(x-4)    ;  a is the multiplier to be found

 

f(x) = a (x + 3)(x + 3)(x - 2) (x - 4)

 

Polynomial contains (1, 96);  substitute into equation

 

96 = a (4)(4)(-1)(-3)

 

96 = 48a

 

48 = a

 

Equation:   f(x) = 2(x+3)(x+3)(x-2)(x-4)