Factor x^2 - 5x + 8?

Answer 1

Use the quadratic equation. Interpret your equation in the form of ax^2 + bx +c then

x= [-b +- √(b^2 - 4ac)]/2a (The quadratic formula)

x= [5 +-√(25-4(1)(8)]/2
x=[5+-√(-7)]/2 or if you prefer [5+-i√(7)]/2)
so the factors are
(x - [5+i√(7)]/2)(x - [5 -i√(7)]/2)

Answer 2

In order to factor this, we need to call it p(x) and set it to 0. So, if
p(x) = x^2 - 5x + 8

We need to find the roots

x^2 - 5x + 8 = 0

To find the roots, we can either use the quadratic formula or complete the square. I'm gonna use Complete The Square despite how long it takes.

x^2 - 5x + 25/4 + 8 = 25/4

(x - 5/2)^2 = 25/4 - 8
(x - 5/2)^2 = 25/4 - 32/4
(x - 5/2)^2 = -7/4

Therefore,

x - 5/2 = +/- sqrt(7)i/2
x = [5 +/- sqrt(7)i]/2
x = 5/2 +/- [sqrt(7)/2]i

Therefore, our two roots are
x = 5/2 + [sqrt(7)/2]i and x = 5/2 - [sqrt(7)/2]i
For now, let's denote the first solution as alpha and the second solution as beta.
Since alpha and beta are roots, (x - alpha) and (x - beta) are factors.

Working them out, the factors we get are:
(x - (5/2 + [sqrt(7)/2]i) ) (x - (5/2 - [sqrt(7)/2]i)

Answer 3

let:
x^2 - 5x + 8 = ax^2 + bx + c

look at sqrt (b^2 - 4ac) = sqrt (25 - 32) = sqrt (-7)

so ....
"can't be factored" if you require that coefficients
of the factors' terms be integers (or rational numbers)


/*******/
NOTE: the "general" solution of quadratic equations
is:

x = ( -b [+or-] sqrt (b^2 - 4ac) ) / 2a

so if it "factors" this will give a good clue as to
what those factors are.

Answer 4

It's not factorizable. The equation x^2 - 5x + 8 = 0 doesn't have two natural roots, because its Delta is negative (b^2 - 4ac = -7 < 0). It has two "complex" roots.

Answer 5

Factorizar?

x(x-5)+8??'

Answer 6

Can't binomial factor it. Use the quadratic formula or complete the square,.

Answer 7

x^2 - 5x +8 = x^2 - 5x + 8. QED