Question

Given that 4-2i is a zero, factor the following polynomial

Answer 1

Step-by-step explanation

We must factor the following polynomial:

`f(x)=x^4-23x^3+196x^2-784x+1120.`

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:

`\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}`

Where a = 4 - 2i, b = 4 + 2i, c and d are roots.

(3) Making the product of the last equation, we get:

`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.`

Comparing the coefficients of x³ and the constant term with the polynomial of the statement, we see that we must have:

`\begin{gathered} -8-c-d=-23\Rightarrow c=-d-8+23=15-d, \\ 20cd=1120\Rightarrow d=(1120)/(20c)=(56)/(c). \end{gathered}`

Replacing the second equation in the first equation, and solving for c, we get:

`c=7.`

Using the equation for d, we get:

`d=8.`

(4) Using the results above, we write the factorized form of the polynomial:

`f(x)=(x-4+2i)(x-4-2i)(x-7)(x-8).`Answer
`f(x)=(x-4+2i)(x-4-2i)(x-7)(x-8)`