How do you factor completely #x^2 - 7x + 30#?

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Answer 1 · 2015-12-22T20:09:36

#x^2-7x+30 = (x-7/2-sqrt(71)/2 i)(x-7/2+sqrt(71)/2 i)#

#x^2-7x+30# is of the form #ax^2+bx+c# with #a=1#, #b=-7# and #c=30#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-7)^2-(4xx1xx30) = 49-120 = -71#

Since this is negative, the quadratic has no Real zeros and no simpler factors with Real coefficients.

If you still want to factor it, you can use the quadratic formula to find the Complex conjugate pair of zeros and hence derive the factors:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (-b+-sqrt(Delta))/(2a)#

#=(7+-sqrt(-71))/2#

#=(7+-i sqrt(71))/2#

#=7/2+-sqrt(71)/2 i#

Hence:

#x^2-7x+30 = (x-7/2-sqrt(71)/2 i)(x-7/2+sqrt(71)/2 i)#