Write a polynomial function f of least degree

Answered question

2022-01-19

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

Answer & Explanation

xleb123

xleb123

Skilled2022-01-30Added 181 answers

We are given a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros 2i,1i.

As we know that the polynomial function has rational coefficients.

so any complex root always appear in pair.

so as 1i is a root of f(x) then it's complex conjugate 1+i is also an root of f(x).

Hence f(x) could be represented as:  

f(x)=(x2i)[x(1i)][x(1+i)]

Simplify each term.

Apply the distributive property.

f(x)=(x2i)(x11+i)(x(1+i))

Multiply −1 by 1.

f(x)=(x2i)(x1+i)(x(1+i))

Multiply −1 by -1.

f(x)=(x2i)(x1+1i)(x(1+i))

Multiply i by 1.

f(x)=(x2i)(x1+i)(x(1+i))

Expand (x2i)(x1+i) by multiplying each term in the first expression by each term in the second expression.

f(x)=(xx+x1+xi2ix2i12ii)(x(1+i))

f(x)=(x2x+xi2ix+2i+2)(x(1+i))

Subtract 2ix from xi.

f(x)=(x2xxi+2i+2)(x(1+i))

Simplify each term.

f(x)=(x2xxi+2i+2)(x1i)

Expand (x2xxi+2i+2)(x1i) by multiplying each term in the first expression by each term in the second expression.

f(x)=x2x+x21+x2(i)xxx1x(i)xixxi1xi(i)+2ix+2i1+2i(i)+2x+21+2(i)

Simplify terms.

f(x)=x32x2+x2(i)+xix2i+xi+2ix2i+2x2i

Subtract x2i from x2(i).

f(x)=x32x22x2i+xi+xi+2ix2i+2x2i

Add xi and xi.

f(x)=x32x22x2i+2xi+2ix2i+2x2i

Add 2xi and 2ix.

f(x)=x32x22x2i+4xi2i+2x2i

Subtract 2i from 2i.

f(x)=x32x22x2i

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