How do you factor the trinomial #2x^2 - 8x + 5#?

1 Answer
George C.
May 15, 2016

#2x^2-8x+5=2(x-2-sqrt(6)/2)(x-2+sqrt(6)/2)#

Explanation:

Complete the square, then use the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(x-2)# and #b=sqrt(6)/2# as follows:

#2x^2-8x+5#

#=2(x^2-4x+5/2)#

#=2(x^2-4x+4-6/4)#

#=2((x-2)^2-(sqrt(6)/2)^2)#

#=2((x-2)-sqrt(6)/2)((x-2)+sqrt(6)/2)#

#=2(x-2-sqrt(6)/2)(x-2+sqrt(6)/2)#

#color(white)()#
Footnote

Why did I choose this method, rather than trying an AC method, etc.?

#2x^2-8x+5# is in the form #ax^2+bx+c# with #a=2#, #b=-8# and #c=5#.

This has discriminant given by the formula:

#Delta = b^2-4ac = (-8)^2-(4*2*5) = 64-40 = 24#

which is not a perfect square, so the factors will not have rational coefficients.

We could use the quadratic formula to find them, but completing the square is just as powerful and less "magic".

Related questions
See all questions in Factorization of Quadratic Expressions
Impact of this question
1198 views around the world
You can reuse this answer
Creative Commons License