Since 3+2i is a solution, then 3-2i must also be a solution if the resulting
polynomial is to have all real coefficients.

Start with these three equations for "x = ↏"

x = -5,  x = 3+2i, and we must also include x = 3-2i

Get 0 on the right of each

x+5 = 0, x-3-2i = 0, x-3+2i = 0

Multiply equals by equals:

(x+5)(x-3-2i)(x-3+2i) = 0

Group the (x-3)'s

(x+5)[(x-3)-2i][(x-3)+2i] = 0

Multiply the last two factors:

(x+5)[(x-3)²-4i²] = 0

(x+5)[(x-3)²-4(-1)] = 0

(x+5)[(x-3)²+4] = 0

(x+5)[(x-3)(x-3)+4] = 0

(x+5)[x²-6x+9+4] = 0

(x+5)[x²-6x+13] = 0

x³-6x²+13x+5x²-30x+65

x³-x²-17x+65 = 0

So the function of least degree with real coefficients 
that has those zeros is

f(x) = x³-x²-17x+65

[Note:  If the problem did not state "with real coefficients", then 
this: f(x) = x²+(2-2i)x-(10i+15), with imaginary coefficients would  
be the polynomial with least degree, but I'm almost sure the problem 
stated "with real coefficients".] 

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Start with these three equations for "x = ↏"

x = -1, x = 5, x = 2

Do the same as above and see if you can get

f(x) = x³-6x²+3x+10

Edwin