How do you find all the zeros of f(x) = x^2 + 6x + 18f(x)=x2+6x+18?

1 Answer
Mar 10, 2016

Complete the square and use the difference of squares identity to find zeros:

x = -3+-3ix=3±3i

Explanation:

Complete the square then use the difference of squares identity:

a^2-b^2 = (a-b)(a+b)a2b2=(ab)(a+b)

with a=(x+3)a=(x+3) and b=3ib=3i as follows:

f(x) = x^2+6x+18f(x)=x2+6x+18

= x^2+6x+9+9=x2+6x+9+9

= (x+3)^2+3^2=(x+3)2+32

=(x+3)^2-(3i)^2=(x+3)2(3i)2

= ((x+3)-3i)((x+3)+3i)=((x+3)3i)((x+3)+3i)

= (x+3-3i)(x+3+3i)=(x+33i)(x+3+3i)

Hence f(x) = 0f(x)=0 when x=-3+-3ix=3±3i