Question 150527
Are you sure that the polynomial is not of degree 3? If it's degree 18, then what is the multiplicity of each root?



So I'm going to assume that the polynomial is of degree 3.



Since {{{3}}}, {{{1+i}}}, and {{{1-i}}} are given zeros this means that:



{{{x=3}}}, {{{x=1+i}}}, and {{{x=1-i}}}



Get all terms to the left side in each case



{{{x-3=0}}}, {{{x-(1+i)=0}}}, and {{{x-(1-i)=0}}}




{{{(x-3)(x-(1+i))(x-(1-i))=0}}} Now use the zero product property in reverse to join the factors.



{{{(x-3)((x-1)+i)((x-1)-i)=0}}} Regroup the terms



{{{(x-3)((x-1)^2-i^2)=0}}} Factor {{{((x-1)+i)((x-1)-i)}}} to get {{{(x-1)^2-i^2}}} by use of the difference of squares. Note: let {{{A=x-1}}} and rewrite the problem into {{{(A+i)(A-i)}}}



{{{(x-3)((x-1)^2-(-1))=0}}} Rewrite {{{i^2}}} as {{{-1}}}



{{{(x-3)((x-1)^2+1)=0}}} Rewrite {{{(x-1)^2-(-1)}}} as {{{(x-1)^2+1}}}



{{{x(x^2-2x+2)-3(x^2-2x+2)}}} Expand. Remember, {{{(a+b)(c+d+e)=a(c+d+e)+b(c+d+e)}}}



{{{(x)*(x^2)+(x)*(-2x)+(x)*(2)+(-3)*(x^2)+(-3)*(-2x)+(-3)*(2)}}} Distribute.



{{{x^3-2*x^2+2*x-3*x^2+6*x-6}}} Multiply.



{{{x^3-5*x^2+8*x-6}}} Now combine like terms.




So the polynomial of degree 3 that has the roots {{{3}}}, {{{1+i}}}, and {{{1-i}}} is {{{x^3-5*x^2+8*x-6}}}




Notice how if we graph {{{y=x^3-5*x^2+8*x-6}}}, we can visually verify our answer (note: in this case, we can only verify the root {{{3}}})


{{{ graph( 500, 500, -10, 10, -10, 10, x^3-5*x^2+8*x-6 ) }}} Graph of {{{y=x^3-5*x^2+8*x-6}}} with root of {{{x=3}}}