How do you find the zeros of #g(x)=x^3+4x^2+7x+28#?

1 Answer
Dec 14, 2016

Answer:

# x=+-sqrt(7)i # or #x=-4#

Explanation:

As this is a cubic there will be three solutions. By the context of the question two of the solutions will be complex (conjugate pairs), and so the remaining solution will be real.

The first task is to factorise the cubic into quadratic and linear factors.

# \ \ \ \ \ g(x)=x^3+4x^2+7x+28 #

Which by observation we can write as:

# \ \ \ \ \ g(x)=x^2(x+4)+7(x+4) #
# :. g(x)=(x^2+7)(x+4) #

To find the roots we set #g(x)=0 =>#:

# (x^2+7)(x+4) = 0#
# :. (x^2+7) = 0# or #(x+4)=0#
# :. x^2=-7# or #x=-4#
# :. x=+-sqrt(7)i# or #x=-4#