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

Question

Eli Assouline · Accepted Answer

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

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".

Jameson Allen · Answer

undefined To factor the trinomial $2x^2 - 8x + 5$, you can use the quadratic formula or factor by grouping. Let's use the latter method:

1. Multiply the leading coefficient (2) by the constant term (5): $2 \times 5 = 10$.
2. Find two numbers that multiply to give 10 and add to give the middle coefficient (-8). These numbers are -2 and -5.
3. Rewrite the middle term (-8x) using the two numbers found in step 2: $2x^2 - 2x - 5x + 5$.
4. Group the terms: $(2x^2 - 2x) + (-5x + 5)$.
5. Factor out the greatest common factor from each group: $2x(x - 1) - 5(x - 1)$.
6. Factor out the common binomial factor $x - 1$: $(2x - 5)(x - 1)$.

Therefore, the factored form of $2x^2 - 8x + 5$ is $(2x - 5)(x - 1)$.