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 xleb123

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 $f\left(x\right)$ then it's complex conjugate $1+i$ is also an root of $f\left(x\right).$

Hence $f\left(x\right)$ could be represented as:

$f\left(x\right)=\left(x-2i\right)\left[x-\left(1-i\right)\right]\left[x-\left(1+i\right)\right]$

Simplify each term.

Apply the distributive property.

$f\left(x\right)=\left(x-2i\right)\left(x-1\cdot 1+i\right)\left(x-\left(1+i\right)\right)$

Multiply −1 by 1.

$f\left(x\right)=\left(x-2i\right)\left(x-1+i\right)\left(x-\left(1+i\right)\right)$

Multiply −1 by -1.

$f\left(x\right)=\left(x-2i\right)\left(x-1+1i\right)\left(x-\left(1+i\right)\right)$

Multiply $i$ by $1$.

$f\left(x\right)=\left(x-2i\right)\left(x-1+i\right)\left(x-\left(1+i\right)\right)$

Expand $\left(x-2i\right)\left(x-1+i\right)$ by multiplying each term in the first expression by each term in the second expression.

$f\left(x\right)=\left(x\cdot x+x\cdot -1+xi-2ix-2i\cdot -1-2ii\right)\left(x-\left(1+i\right)\right)$

$f\left(x\right)=\left({x}^{2}-x+xi-2ix+2i+2\right)\left(x-\left(1+i\right)\right)$

Subtract $2ix$ from $xi.$

$f\left(x\right)=\left({x}^{2}-x-xi+2i+2\right)\left(x-\left(1+i\right)\right)$

Simplify each term.

$f\left(x\right)=\left({x}^{2}-x-xi+2i+2\right)\left(x-1-i\right)$

Expand $\left({x}^{2}-x-xi+2i+2\right)\left(x-1-i\right)$ by multiplying each term in the first expression by each term in the second expression.

$f\left(x\right)={x}^{2}x+{x}^{2}\cdot -1+x2\left(-i\right)-x\cdot x-x\cdot -1-x\left(-i\right)-xix-xi\cdot -1-xi\left(-i\right)+2ix+2i\cdot -1+2i\left(-i\right)+2x+2\cdot -1+2\left(-i\right)$

Simplify terms.

$f\left(x\right)={x}^{3}-2{x}^{2}+{x}^{2}\left(-i\right)+xi-{x}^{2}i+xi+2ix-2i+2x-2i$

Subtract ${x}^{2}i$ from ${x}^{2}\left(-i\right).$

$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i+xi+xi+2ix-2i+2x-2i$

Add $xi$ and $xi.$

$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i+2xi+2ix-2i+2x-2i$

Add $2xi$ and $2ix.$

$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i+4xi-2i+2x-2i$

Subtract $2i$ from $-2i.$

$f\left(x\right)={x}^{3}-2{x}^{2}-2{x}^{2}i$

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