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)a2−b2=(a−b)(a+b)
with
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+3−3i)(x+3+3i)
Hence