Answer:
The polynomial of minimum degree is [tex]p(x) = x^{4}-3\cdot x^{3}-44\cdot x^{2}+292\cdot x-480[/tex].
Step-by-step explanation:
According to the statement, we appreciate that polynomial pass through the following points:
(i) [tex](4 + 2\,i,0)[/tex], (ii) [tex](4 - 2\,i, 0)[/tex], (iii) [tex](-8, 0)[/tex], (iv) [tex](0, -480)[/tex]
From Algebra we know that any n-th grade polynomial can be constructed by knowing n+1 different points. Hence, the polynomial of minimum degree is a quartic function. The polynomial has the following form:
[tex]p(x) = (x-4-2\,i)\cdot (x-4+2\,i)\cdot (x+8)\cdot (x-r_{1})[/tex]
We proceed to expand the expression until standard form is obtained:
[tex]p(x) = (x^{2}-4\cdot x-2\,i\cdot x-4\cdot x +16+8\,i+2\,i\cdot x-8\,i+4)\cdot (x+8)\cdot (x-r_{1})[/tex]
[tex]p(x) = (x^{2}-8\cdot x+20)\cdot (x+8)\cdot (x-r_{1})[/tex]
[tex]p(x) = (x^{3}-8\cdot x^{2}+20\cdot x +8\cdot x^{2}-64\cdot x+160)\cdot (x-r_{1})[/tex]
[tex]p(x) = (x^{3}-44\cdot x +160)\cdot (x-r_{1})[/tex]
If we know that [tex]p (0) = -480[/tex], then:
[tex]-480=160\cdot (-r_{1})[/tex]
[tex]-160\cdot r_{1} = -480[/tex]
[tex]r_{1} = 3[/tex]
Then, the polynomial is:
[tex]p(x) = (x^{3}-44\cdot x +160)\cdot (x-3)[/tex]
[tex]p(x) = x^{4}-44\cdot x^{2}+160\cdot x-3\cdot x^{3}+132\cdot x -480[/tex]
[tex]p(x) = x^{4}-3\cdot x^{3}-44\cdot x^{2}+292\cdot x-480[/tex]
The polynomial of minimum degree is [tex]p(x) = x^{4}-3\cdot x^{3}-44\cdot x^{2}+292\cdot x-480[/tex].
