How do you factor # x^2-3x+15#?

Question

5300 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-08T12:27:46

Examine to discriminant to find that this has no factors with Real coefficients. Use the quadratic formula to find:

#x^2-3x+15 = (x-3/2-isqrt(51)/2)(x-3/2+isqrt(51)/2)#

#x^2-3x+15# is of the form #ax^2+bx+c# with #a=1#, #b=-3# and #c=15#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-3)^2-(4xx1xx15) = 9 - 60 = -51#

Since #Delta < 0#, #x^2-3x+15 = 0# has no Real roots and #x^2-3x+15# has no factors with Real coefficients.

The complete conjugate roots of #x^2-3x+15 = 0# are given by the quadratic formula:

#x = (-b+-sqrt(Delta))/(2a) = (3+-isqrt(51))/2#

So #x^2-3x+15 = (x-3/2-isqrt(51)/2)(x-3/2+isqrt(51)/2)#