Question

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

Answer 1

`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)`

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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".