A polynomial function of least degree with integral coefficients that has the
given zeros 
Given
Given zeros are 3i, -1 and 0
complex zeros occurs in pairs. 3i is one of the zero
-3i is the other zero
So zeros are 3i, -3i, 0 and -1
Now we write the zeros in factor form
If 'a' is a zero then (x-a) is a factor
the factor form of given zeros
Now we multiply it to get the polynomial
polynomial function of least degree with integral coefficients that has the
given zeros
Learn more : brainly.com/question/7619478
Answer:
Some of the prime factors are 1,2,4,5,10,20,-1,-.2,-4,-5,-10,-20, and since 5 is prime you can just start with 5 and 4
Step-by-step explanation:
You cant simplify it . it is already simplified
Answer:
Vertex form: f(x) = -10(x − 2)^2 + 3
Standard form: y = -10x^2 + 40x - 37
Step-by-step explanation:
h and k are the vertex coordinates
Substitute them in the vertex form equation:
f(x) = a(x − 2)^2 + 3
Calculate "a" by replacing "f(x)" with -7 and "x" with 1:
-7 = a(1 − 2)^2 + 3
Simplify:
-7 = a(1 − 2)^2 + 3
-7 = a(-1)^2 + 3
-7 = a + 3
-10 = a
Replace a to get the final vertex form equation:
f(x) = -10(x − 2)^2 + 3
Convert to standard form:
y = -10(x − 2)^2 + 3
Expand using binomial theorem:
y = -10(x^2 − 4x + 4) + 3
Simplify:
y = -10x^2 + 40x - 40 + 3
y = -10x^2 + 40x - 37
