Question

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

Answer 1

Answer:

The simplest polynomial function is $y = x^{4}-(7+\sqrt{13})\cdot x^{3}+(27+7\sqrt{13})\cdot x^{2}-(189+27\sqrt{13})\cdot x + 189\sqrt{13}$.

Step-by-step 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:

$y = (x-7)\cdot (x-\sqrt{13})\cdot (x-i\,27)\cdot (x+i\,27)$

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

$y = [x^{2}-(7+\sqrt{13})\cdot x +7\sqrt{13}]\cdot (x^{2}+27)$

$y = x^{2}\cdot (x^{2}+27)-(7+\sqrt{13})\cdot [(x^{2}+27)]\cdot x+7\sqrt{13}\cdot (x^{2}+27)$

$y = x^{4}+27\cdot x^{2}-(7+\sqrt{13})\cdot x^{3}-(7+\sqrt{13})\cdot 27\cdot x +7\sqrt{13}\cdot x^{2}+189\sqrt{13}$

$y = x^{4}-(7+\sqrt{13})\cdot x^{3}+(27+7\sqrt{13})\cdot x^{2}-(189+27\sqrt{13})\cdot x + 189\sqrt{13}$

The simplest polynomial function is $y = x^{4}-(7+\sqrt{13})\cdot x^{3}+(27+7\sqrt{13})\cdot x^{2}-(189+27\sqrt{13})\cdot x + 189\sqrt{13}$.