Form a​ third-degree polynomial function with real​ coefficients, with leading coefficient​ 1, such that 6+i and 5 are zeros.

A third-degree polynomial can be written as:

f(x) = a*x^3 + b*x^2 + c*x + d

Where the leading coefficient is a, and all the coefficients are real.

If we know that the leading coefficient is 1, then the equation becomes:

f(x) = x^3 + b*x^2 + c*x + d

Now, we also know that:

(6 + i) and 5 are zeros.

This means that:

(6 + i)^3 + b*(6 + i)^2 + c*(6 + i) + d = 0

remember that:

i^2 = - 1

This is equal to:

(6 + i)*(36 + 2*6*i  + i^2) + b*(36 + 2*6*i  + i^2) + c*(6 + i) + d = 0

(6 + i)*(35 + 12i) + b*(35 + 12i) + c*(6 + i) + d  =0

(210 + 35i + 72i - 12) +  b*(35 + 12i) + c*(6 + i) + d  = 0

198 + 107i +  b*(35 + 12i) + c*(6 + i) + d  = 0

sparating in real and imaginary part, we get:

(198 + b*35 + c*6 + d) + (107 + b*12 + c)*i = 0

Then each parentheses needs to be zero, this means that:

198 + b*35 + c*6 + d = 0

107 + b*12 + c =  0

Knowing that 5 is another zero, we have:

5^3 + b*5^2 + c*5 + d = 0

125 + b*25 + c*5 + d = 0

Then we have a system of 3 equations and 3 variables:

198 + b*35 + c*6 + d = 0

107 + b*12 + c =  0

125 + b*25 + c*5 + d = 0

To solve this, we first need to isolate one of the variables in one of the equations.

Let's isolate d in the last one, so we get:

d = -125 - b*25 - c*5

now we can replace this in the first equation to get:

198 + b*35 + c*6 + d = 0

198 + b*35 + c*6 + (  -125 - b*25 - c*5) = 0

70 + b*10 + c = 0

So now we have two equations:

70 + b*10 + c = 0

107 + b*12 + c =  0

Again, now we can isolate the one variable in one of the equations, this time let's isolate c in the first one.

c = -70 - b*10

now we can replace this in the other equation:

107 + b*12 + c = 0

107 + b*12 + (-70 - b*10) = 0

38 + b*2 = 0

now we can solve this for b

b*2 = -38

b = -38/2 = -19

Now, with the equation c = -70 - b*10 we can find the value of c.

c = -70 - b*10 = c = -70 - (-19)*10 = 120

And with the equation  d = -125 - b*25 - c*5

we can find the value of d:

d = -125  - b*25 - c*5 = -125 - (-19)*25 - (120)*5 = -250

Then we have:

a = 1

b = -19

c = 120

d = -250

The eqation is:

f(x) = 1*x^3 - 19*x^2 + 120*x - 250