How do you find the zeros of #f(x)=x^3+3x^2+6x+4#?

1 Answer
Jan 17, 2017

Answer:

#f(x)# has zeros #-1# and #-1+-sqrt(3)i#

Explanation:

Given:

#f(x) = x^3+3x^2+6x+4#

Note that #-1+3-6+4 = 0#, so #f(-1) = 0#, #x=-1# is a zero and #(x+1)# a factor:

#x^3+3x^2+6x+4 = (x+1)(x^2+2x+4)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x+1)# and #b=sqrt(3)i# as follows:

#x^2+2x+4 = x^2+2x+1+3#

#color(white)(x^2+2x+4) = (x+1)^2-(sqrt(3)i)^2#

#color(white)(x^2+2x+4) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)#

#color(white)(x^2+2x+4) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)#

Hence the other two zeros are:

#x = -1+-sqrt(3)i#