How do you factor $2x^2 - 8x + 5$ ?

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Answer 1 · 2015-05-24T11:13:15

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

This has discriminant given by the formula:

$Delta = b^2-4ac = (-8)^2 - (4xx2xx5) = 64 - 40 = 24$

$Delta > 0$ implies that $2x^2-8x+5 = 0$ has two distinct real roots and therefore two linear factors with real coefficients.

Unfortunately, $Delta$ is not a perfect square, so those coefficients are not rational.

The roots of $2x^2-8x+5 = 0$ are given by the formula:

$x = (-b+-sqrt(Delta))/(2a) = (8 +- sqrt(24))/4$

$= 2+- sqrt(2^2*6)/4 = 2+-(2sqrt(6))/4 = (4+-sqrt(6))/2$

For symmetry I would like to multiply both sides of this by $sqrt(2)$ to get:

$sqrt(2)x = (4sqrt(2)+-sqrt(2)sqrt(6))/2$

$= (4sqrt(2)+-sqrt(2)sqrt(2)sqrt(3))/2$

$= (4sqrt(2)+-2sqrt(3))/2 = 2sqrt(2)+-sqrt(3)$

and we find:

$2x^2-8x+5 = (sqrt(2)x-2sqrt(2)+sqrt(3))(sqrt(2)x-2sqrt(2)-sqrt(3))$

How do you factor 2x^2 - 8x + 52x^2 - 8x + 5?