Write the complex number in the form a + bi.

square root of six(cos 315° + i sin 315°)

Step-by-step explanation:

The given complex number is,

we know that,

and

Putting the values,

Quadratic formula:

x⁴ - 14x² + 32

x = - (-14) ± √(-14)² - 4(1)(32) /2(1)

x = 14 ± √196 - 128 / 2

x = 14 ± √68 / 2

x = 14 ± 2√17 / 2

Factor the numerator

x = 2(7 ± √17) / 2

Cancel out the two on both the numerator and denominator.

x = 7 ± √17

The two zeros for this problem are 7 ± √17.

So far, the zeros are irrational. These roots are real, unequal, irrational roots. The zeros are not perfect squares.

If you graph this polynomial function into a graphing calculator, you'll see that the parabola doesn't intersect the x-axis at rational values.

Furthermore, I have tried to factor this polynomial function by doing the AC method and grouping. However, the closest factorization I got was:

(x² - 12)(x² - 2.667) which is not even equivalent to the given function.

Although this function has real zeros, I believe that it is impossible to factor. Therefore, my answer is that this function is not factorable.