Question

Given that 3-3i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable.

f(x) = x² + 5x³ - 20x² + 30x +504
4
Answer
f(x) =

Answer 1

Factoring the polynomial (f(x)) with the given zero (3 - 3i) and applying the Conjugate Roots Theorem yields the completely factored form
`\(f(x) = (x - 3)^2(x^2 + 9)\)`.

To factor the given polynomial
`\(f(x) = x^2 + 5x^3 - 20x^2 + 30x + 504\)`, it is mentioned that (3 - 3i) is a zero. According to the Conjugate Roots Theorem, if a complex number (a + bi) is a zero of a polynomial with real coefficients, then its conjugate (a - bi) must also be a zero. Therefore, if (3 - 3i) is a zero, then (3 + 3i) is also a zero.

Now, we can use these zeros to factor (f(x)). The factored form is given by:

`\[ f(x) = a(x - r_1)(x - r_2)(x - r_3)\ldots(x - r_n) \]`

where \(a\) is the leading coefficient, and
`\(r_1, r_2, \ldots, r_n\)` are the zeros.

Since (3 - 3i) and (3 + 3i) are zeros, the factors involving these roots are:

`\[ (x - (3 - 3i))(x - (3 + 3i)) \]`

Now, we can expand and simplify this expression:

`\[ (x - 3 + 3i)(x - 3 - 3i) \]`

Using the difference of squares formula
`(\(a^2 - b^2 = (a - b)(a + b)\))`, we get:

`\[ (x - 3)^2 - (3i)^2 \]`

`\[ (x - 3)^2 + 9 \]`

So, the factored form of the polynomial (f(x)) is:

`\[ f(x) = a(x - r_1)(x - r_2)(x - r_3)\ldots(x - r_n) \]`

`\[ f(x) = (x - 3)^2(x^2 + 9) \]`