Explanation
We must factor the following polynomial:
[tex]f(x)=x^4-23x^3+196x^2-784x+1120.[/tex]
We must use the fact that 4 - 2i is a zero and the Complex Conjugate Root Theorem.
(1) The Complex Conjugate Root Theorem states that if P is a polynomial in one variable with real coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a − bi is also a root of P.
Knowing that a = 4 - 2i is a root, and using this theorem, we conclude that b = 4 + 2i is also a root of f(x).
(2) The factorize form of f(x) is:
[tex]\begin{gathered} f(x)=1\cdot(x-a)(x-b)(x-c)(x-d), \\ =(x-4+2i)(x-4-2i)(x-c)(x-d). \end{gathered}[/tex]
Where a = 4 - 2i, b = 4 + 2i, c and d are roots.
(3) Making the product of the last equation, we get:
[tex]f(x)=1\cdot x^4+(-8-c-d)\cdot x^3+(20+8c+8d+cd)\cdot x^2+(-20c-20d-8cd)\cdot x+20cd.[/tex]
Comparing the coefficients of x³ and the constant term with the polynomial of the statement, we see that we must have:
[tex]\begin{gathered} -8-c-d=-23\Rightarrow c=-d-8+23=15-d, \\ 20cd=1120\Rightarrow d=\frac{1120}{20c}=\frac{56}{c}. \end{gathered}[/tex]
Replacing the second equation in the first equation, and solving for c, we get:
[tex]c=7.[/tex]
Using the equation for d, we get:
[tex]d=8.[/tex]
(4) Using the results above, we write the factorized form of the polynomial:
[tex]f(x)=(x-4+2i)(x-4-2i)(x-7)(x-8).[/tex]Answer[tex]f(x)=(x-4+2i)(x-4-2i)(x-7)(x-8)[/tex]
