how do i factor.....?

Question

factor

x^2 - 3x - 12

&

x^2 - x - 10

Answer 1

Completing the Square

x² - 3x - 12 = 0

x² - 3x - 12 + 12 = 0 + 12

x² - 3x = 12

x² - 3x + ________ = 12 + _______

x² - 3x + 9/4 = 12 + 9/4

(x - 3/2)(x - 3/2) = 48/4 + 9/4

(x - 3/2)² = 57/4

(√x - 3/2) = ± √57 / √4

x - 3/2 = ± √57 / 2

x - 3/2 = ± 7.549834435 / 2

x - 3/2 + 3/2 = 3/2 ± 7.549834435 / 2

x = 3/2 ± 7.549834435 / 2

- - - - - - - - - -

Solving for +

x = 3/2 + 7.549834435 / 2

x = 10.54983444 / 2

x = 5.274917218

- - - - - - - - - -

Solving for -

x = 3/2 - 7.549834435 / 2

x = - 4.549834435 / 2

x = - 2.274917218

- - - - - - - - - - - - - - -

Quadratic formula also works

x = - b ± √b² - 4ac / 2a

- - - - - - - - -s-

Answer 2

Neither of them will factor 'nicely', but you can solve them for the roots using the quadratic formula
x = (-b ± √(b²-4ac))/2a
which gives x values at which ax²+bx+c=0.

This will give you 2 values for x (call them p and q) and the trinomial then factors as
(x-p)(x-q)

HTH

Doug

Answer 3

use completing square method i.e.
x^2-3x-12
[x^2-3x+(3/2)^2]-(3/2)^2-12
[(x-3/2)^2]-9/4-12
[(x-3/2)^2]-57/4
(x-3/2)^2-[sqr(57/4)]^2
[x-3/2+sqr(57/4)][x-3/2-sqr(57/4)]
(using a^2-b^2=(a+b)(a-b))
now these are your required factors
like this you can also factorise second one.

Answer 4

Neither of these are factorable with integers.