Find a polynomial equation with real coefficients that has
the roots of -5, 2+i
Since 2+i is a solution, and since the coefficients are all real,
its conjugate 2-i is also a solution. Begin with
x = 5; x = 2+i; x = 2-i
Get 0 alone on the right side of each:
x+5 = 0; x-2-i = 0; x-2+i = 0
Multiply all left sides together [and right sides too! (0)(0)(0)=0]
(x+5)(x-2-i)(x-2+i) = 0
Simplify:
(x+5)[(x-2)-i][(x-2)+i] = 0
(x+5)[(x-2)²-i²] = 0
(x+5)[(x-2)²-(-1)] = 0
(x+5)[(x-2)²+1] = 0
(x+5)[(x-2)(x-2)+1] = 0
(x+5)[x²-4x+4+1] = 0
(x+5)(x²-4x+5) = 0
x³+x²-15x+25 = 0
So those zeros would be obtained when we set the
function f(x) = x³+x²-15x+25 equal to 0. So
one answer is:
f(x) = x³+x²-15x-25
Other answers would be gotten by multiplying the
right side by different constants k:
f(x) = kx³+kx²+25kx-15k
Edwin