Polynomial Function Given Integer Zeros

Question

Here's the question:

Let f(x) be a cubic polynomial with zeros -1, 2, and 3. If the graph of f passes through (0, 3) write the completely factored form of f(x).

 

So, what I have is this: I took the zeros and plugged them into this equation: (x+1)(x-2)(x-3) and got x^3-4x^2+x+6 but when I graph that, I'm confused because it doesn't pass through (0,3) so am I starting it wrong or am I not understanding where to go from here?

Answer 1

Knowing zeros of a polynomial is only sufficient to write a polynomial expression if its numerical coefficient "a" is equal to "1".

It's not the case here.

The general form of this polynomial is

y = a(x + 1)(x - 2)(x - 3)

Substituting x, y values for point (0, 3):

3 = a(0 + 1)(0 - 2)(0 - 3) = a(1•(-2)•(-3)) = 6a

6a = 3

a =1/2

Therefore, the completely factored polynomial is

y = (1/2)(x +1)(x - 2)(x - 3)

Answer 2

Your equation should be y =a(x+1)(x-2)(x-3).

you can construct an infinite number of graphs with zeros at -1,2and 3 The one you want depends on the leading coefficient (a)

you can solve for a by using the given point (0,3) to solve for a

3=a(0+1)(0-2)(0-3)

3=a(6)

a=1/2

so equation is y= (1/2)( x+1)(x-2)(x-3)

graph this in desmos and you will see this satisfies the stated parameters

Then change the a value and graph and you will see you still have a graph with the same zeroes but every thing else is different