Question

The complex zeros of the following polynomial function. Write f in factored form.

Answer 1

Answer:  $\bold{x=-3\qquad x=\dfrac{3+3i\sqrt3}{2}\qquad x=\dfrac{3-3i\sqrt3}{2}}$

Step-by-step explanation:

The formula for factoring a cubic is:

a³ + b³ = (a + b)(a² - ab + b²)

f(x) = x³ + 3³

     = (x + 3)(x² - 3x + 9)

To find the zeros, set each factor equal to zero and solve.

0 = x + 3      0 = x² - 3x + 9

-3 = x                use quadratic formula

$x^2-3x+9\quad \rightarrow \quad a=1, b=-3, c=9\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(9)}}{2(1)}\\\\\\x=\dfrac{3\pm\sqrt{9-36}}{2}\\\\\\x=\dfrac{3\pm\sqrt{-27}}{2}\\\\\\x=\dfrac{3\pm3i\sqrt{3}}{2}$

Answer:

$x=-3\qquad x=\dfrac{3+3i\sqrt3}{2}\qquad x=\dfrac{3-3i\sqrt3}{2}$