So, the cubic polynomial function is 
No, none of the roots have multiplicity.
Step-by-step explanation:
We need to write an equation for the cubic polynomial function
whose graph has zeroes at 2, 3, and 5.
Zeros mean:
x=2, x=3 and x=5
or
x-2=0, x-3=0 and x-5=0
Multiplying all the factors:
So, the cubic polynomial function is
No, none of the roots have multiplicity, A root has multiplicity if it appears more than 1 time. Like if (x-1)^2 is a root then 1 has multiplicity 2
Keywords: Polynomial Function
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X would equal 57 if 52 replaces x then 3+52 is 55 and 55+2 equals 57.
- Vertex/General Form: y = a(x - h)^2 + k, with (h,k) as the vertex
- (x + y)^2 = x^2 + 2xy + y^2
- Standard Form: y = ax^2 + bx + c
So before I put the equation into standard form, I'm first going to be putting it into vertex form. Since the vertex appears to be (-1,7), plug that into the vertex form formula:
Next, we need to solve for a. Looking at this graph, another point that is in this line is the y-intercept (0,5). Plug (0,5) into the x and y placeholders and solve for a as such:
Now we know that <u>our vertex form equation is y = -2(x + 1)^2 + 7.</u>
However, we need to convert this into standard form still, and we can do it as such:
Firstly, solve the exponent: 
Next, foil -2(x^2+2x+1): 
Next, combine like terms and <u>your final answer will be: 
</u>
