Question

n=3; -4 and 3 + 3i are zeros
f(x)= ?

Answer 1

Answer: $\( f(x) = x^3 - 2x^2 - 6x + 72 \)$

Step-by-step explanation:

If \( n = 3 \) and \(-4\) and \(3 + 3i\) are zeros of the function, then the function can be written in factored form using these zeros. Remember, complex zeros always come in conjugate pairs, so if \(3 + 3i\) is a zero, then \(3 - 3i\) must also be a zero.

So, the factored form of the function would be:

$\[ f(x) = a(x - (-4))(x - (3 + 3i))(x - (3 - 3i)) \]$

$\[ f(x) = a(x + 4)(x - 3 - 3i)(x - 3 + 3i) \]$

Now, to find the value of \( a \), we use the given value of \( n \), which is 3. Since the degree of the polynomial is 3, \( a \) is the leading coefficient.

$\[ a = \frac{1}{a}(x + 4)(x - 3 - 3i)(x - 3 + 3i) \]$

$\[ a = \frac{1}{a}(x + 4)((x - 3)^2 - (3i)^2) \]$

$\[ a = \frac{1}{a}(x + 4)((x - 3)^2 - 9i^2) \]$

$\[ a = \frac{1}{a}(x + 4)((x - 3)^2 + 9) \]$

$\[ a^2 = 1 \]$

$\[ a = \pm 1 \]$

Since the leading coefficient can't be negative (it would change the direction of the polynomial's ends), we choose $\( a = 1 \)$.

$So, the function \( f(x) \) is:$

$\[ f(x) = (x + 4)(x - 3 - 3i)(x - 3 + 3i) \]$

$\[ f(x) = (x + 4)(x^2 - 6x + 9 + 9) \]$

$\[ f(x) = (x + 4)(x^2 - 6x + 18) \]$

$\[ f(x) = x^3 - 6x^2 + 18x + 4x^2 - 24x + 72 \]$

$\[ f(x) = x^3 - 2x^2 - 6x + 72 \]$

So, $\( f(x) = x^3 - 2x^2 - 6x + 72 \)$.