Find all complex solutions (real and non-real)

Question

x³ + 12x² + 46x + 60 = 0

Answer 1

Two ways to factor this.

1) By substituting 12x² = 6x² + 6x² and 46x = 10x + 36x then grouping together terms to produce a common factor of (x² + 6x +10)

x³ + 12x² + 46x + 60 = (x³ + 6x² + 10x) + (6x² + 36x +60) = x(x² + 6x +10) + 6(x² + 6x +10)

So x³ + 12x² + 46x + 60 = (x + 6)(x² + 6x +10) The quadratic factor has a negative discriminant (b² – 4ac) so its roots will complex (non-real).

2) By finding the real zero x = –6 by trial-and-error or by graphing the function and seeing that it has an x-intercept at (–6, 0). Once you find the first zero you can use long division come up with

x³ + 12x² + 46x + 60 = (x + 6)(x² + 6x + 10).

You find the two complex zeroes of x² + 6x +10 by using the quadratic formula

x = [–b ±√(b² – 4ac)]2a where a = 1, b = 6, and c = 10.

Answer 2

x^3 +12x^2 +46x +60 has what roots or zeros??

graph it and there's only one x intercept x =-6 = the only real zero

divide x+6 into the 3D polynomial to get a quadratic, then use the quadratic formula to find the conjugate pair of imaginary zeros

3D/(x+6) = x^2 +6x+10

x = -6/2 + or - .5sqr(36 -40) = -3 +/-i

zeros are'

-6, -3+i, -3-i