Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. Write the polynomial in standard form. 2i, 1-i

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We are given a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros 2i,1−i.  As we know that the polynomial function has rational coefficients.  so any complex root always appear in pair.  so as 1−i is a root of f(x) then it&#039;s complex conjugate 1+i is also an root of f(x).  Hence f(x) could be represented as:    f(x)=(x−2i)[x−(1−i)][x−(1+i)]  Simplify each term.  Apply the distributive property.  f(x)=(x−2i)(x−1⋅1+i)(x−(1+i))  Multiply −1 by 1.  f(x)=(x−2i)(x−1+i)(x−(1+i))  Multiply −1 by -1.  f(x)=(x−2i)(x−1+1i)(x−(1+i))  Multiply i by 1.  f(x)=(x−2i)(x−1+i)(x−(1+i))  Expand (x−2i)(x−1+i) by multiplying each term in the first expression by each term in the second expression.  f(x)=(x⋅x+x⋅−1+xi−2ix−2i⋅−1−2ii)(x−(1+i))  f(x)=(x2−x+xi−2ix+2i+2)(x−(1+i))  Subtract 2ix from xi.  f(x)=(x2−x−xi+2i+2)(x−(1+i))  Simplify each term.  f(x)=(x2−x−xi+2i+2)(x−1−i)  Expand (x2−x−xi+2i+2)(x−1−i) by multiplying each term in the first expression by each term in the second expression.  f(x)=x2x+x2⋅−1+x2(−i)−x⋅x−x⋅−1−x(−i)−xix−xi⋅−1−xi(−i)+2ix+2i⋅−1+2i(−i)+2x+2⋅−1+2(−i)  Simplify terms.  f(x)=x3−2x2+x2(−i)+xi−x2i+xi+2ix−2i+2x−2i  Subtract x2i from x2(−i).  f(x)=x3−2x2−2x2i+xi+xi+2ix−2i+2x−2i  Add xi and xi.  f(x)=x3−2x2−2x2i+2xi+2ix−2i+2x−2i  Add 2xi and 2ix.  f(x)=x3−2x2−2x2i+4xi−2i+2x−2i  Subtract 2i from −2i.  f(x)=x3−2x2−2x2i

Write a polynomial function f of least degree

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