Question

Find a polynomial function with integer coefficients that has given zeros. (There are many correct answers)
0, 0, 4, 1+i

Answer 1

When you have two roots that are the same, the this is a double root. This means that x^2 = Double Root
\[x ^{2}=0\]When there is only one real number that is not duplicated, then this is just an x= .\[x=4\]The same goes for the imaginary solutions.\[x=1+i\]
Then you need to make sure that each equation is set equal to zero. This will require moving things back toward the 'x'. This will be like the opposite of solving.
\[x ^{2}=0\]\[x-4=0\]These two are straight forward. The imaginary one will require some work.\[x-1=\sqrt{-1}\]\[(x-1)^{2}=(\sqrt{-1})^{2}\]\[x ^{2}-2x+1=-1\]\[x ^{2}-2x+2=0\]
So, now that the factors are found, just multiply them altogether and this will give you the polynomial.
\[(x ^{2})(x-4)(x ^{2}-2x+2)\]\[(x ^{2})(x ^{3}-6x ^{2}+10x-8)\]\[x ^{5}-6x ^{4}+10x ^{3}-8x ^{2}\]