Question

The simplest polynomial function with zeros 7, square root 13 and 27i would be of what degree

Answer 1

The simplest polynomial function is
`y = x^(4)-(7+√(13))\cdot x^(3)+(27+7√(13))\cdot x^(2)-(189+27√(13))\cdot x + 189√(13)`.

Explanation:

A n-th order polynomial in factorized form is defined by:

`y = (x-r_(1))\cdot (x-r_(2))\cdot ...\cdot (x-r_(n-1))\cdot (x-r_(n))` (1)

Where:

`x` - Independent variable.

`y` - Dependent variable.

`r_(1)`,
`r_(2)`,...,
`r_(n-1)`,
`r_(n)` - Roots of the polynomial.

We know that three roots, two real and a complex root. By Quadratic Formula, a second order polynomial has two conjugated complex number of the form
`a + i\,b` and
`a - i\,b`. Hence, there is an additional zero:
`-i\,27`.

By applying (1), we have the following polynomial when all roots are used:

`y = (x-7)\cdot (x-√(13))\cdot (x-i\,27)\cdot (x+i\,27)`

By factorization and algebraic handling we have the simplest polynomial function:

`y = [x^(2)-(7+√(13))\cdot x +7√(13)]\cdot (x^(2)+27)`

`y = x^(2)\cdot (x^(2)+27)-(7+√(13))\cdot [(x^(2)+27)]\cdot x+7√(13)\cdot (x^(2)+27)`

`y = x^(4)+27\cdot x^(2)-(7+√(13))\cdot x^(3)-(7+√(13))\cdot 27\cdot x +7√(13)\cdot x^(2)+189√(13)`

`y = x^(4)-(7+√(13))\cdot x^(3)+(27+7√(13))\cdot x^(2)-(189+27√(13))\cdot x + 189√(13)`

The simplest polynomial function is
`y = x^(4)-(7+√(13))\cdot x^(3)+(27+7√(13))\cdot x^(2)-(189+27√(13))\cdot x + 189√(13)`.