How do you factor #5x^2 + 30x + 50#?

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Answer 1 · 2016-04-26T21:04:51

#5x^2+30x+50 = 5(x^2+6x+10) =5(x+3-i)(x+3+i)#

Notice that all of the terms are divisible by #5#, so separate that out as a factor first:

#5x^2+30x+50 = 5(x^2+6x+10)#

The quadratic factor #x^2+6x+10# is in the form #ax^2+bx+c# with #a=1#, #b=6# and #c=10#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 6^2-(4*1*10) = 36 - 40 = -4#

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

We can factor it with Complex coefficients, which we can do by completing the square and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with #A=(x+3)# and #B=i# as follows:

#x^2+6x+10#

#=(x+3)^2-9+10#

#=(x+3)^2+1#

#=(x+3)^2-i^2#

#=((x+3)-i)((x+3)+i)#

#=(x+3-i)(x+3+i)#

So putting it all together:

#5x^2+30x+50 = 5(x+3-i)(x+3+i)#