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

1 Answer
George C.
Feb 22, 2016

Use the quadratic formula to find:

#x^2-x+7 = (x-1/2-(3sqrt(3))/2 i)(x-1/2+(3sqrt(3))/2 i)#

Explanation:

#f(x) = x^2-x+7# is in the form #ax^2+bx+c# with #a=1#, #b=-1# and #c=7#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-1)^2-(4*1*7) = 1-28 = -27#

Since this is negative #f(x)# has no linear factors with Real coefficients. We can find its Complex factorisation by using the quadratic formula then converting the zeros into factors:

The roots of #f(x) = 0# are given by the quadratic formula:

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

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

#= (1+-sqrt(-27))/2#

#= 1/2 +- sqrt(27)/2 i#

#= 1/2 +- (3sqrt(3))/2 i#

Hence #f(x)# can be factored as:

#x^2-x+7 = (x-1/2-(3sqrt(3))/2 i)(x-1/2+(3sqrt(3))/2 i)#

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