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

Sagot :

Answer:

The simplest polynomial function is [tex]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}[/tex].

Step-by-step explanation:

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

[tex]y = (x-r_{1})\cdot (x-r_{2})\cdot ...\cdot (x-r_{n-1})\cdot (x-r_{n})[/tex] (1)

Where:

[tex]x[/tex] - Independent variable.

[tex]y[/tex] - Dependent variable.

[tex]r_{1}[/tex], [tex]r_{2}[/tex],..., [tex]r_{n-1}[/tex], [tex]r_{n}[/tex] - 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 [tex]a + i\,b[/tex] and [tex]a - i\,b[/tex]. Hence, there is an additional zero: [tex]-i\,27[/tex].

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

[tex]y = (x-7)\cdot (x-\sqrt{13})\cdot (x-i\,27)\cdot (x+i\,27)[/tex]

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

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

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

[tex]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}[/tex]

[tex]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}[/tex]

The simplest polynomial function is [tex]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}[/tex].