How do you factor the expression #x^2 + 2x + 18#?
1 Answer
Use the quadratic formula to find:
#x^2+2x+18 = (x+1-sqrt(17)i)(x+1+sqrt(17)i)#
Explanation:
Quadratic Formula Method
This has discriminant
#Delta = b^2-4ac = 2^2-(4*1*18) = 4-72 = -68 = -2^2*17#
Since this is negative, this quadratic expression has no Real zeros and no linear factors with Real coefficients.
It can be factored with Complex coefficients.
The zeros are given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#=(-b+-sqrt(Delta))/(2a)#
#=(-2+-sqrt(-68))/2#
#=(-2+-sqrt(68)i)/2#
#=(-2+-2sqrt(17)i)/2#
#=-1+-sqrt(17)i#
Hence the quadratic can be factored:
#x^2+2x+18 = (x+1-sqrt(17)i)(x+1+sqrt(17)i)#
Alternative Method
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
From this we can deduce a "sum of squares" identity:
#a^2+b^2 = a^2-(bi)^2 = (a-bi)(a+bi)#
Then completing the square we find:
#x^2+2x+18#
#=x^2+2x+1+17#
#=(x+1)^2+(sqrt(17))^2#
#=((x+1)-sqrt(17)i)((x+1)+sqrt(17)i)#
#=(x+1-sqrt(17)i)(x+1+sqrt(17)i)#
