factorization

Question

6t² - 7t + 12

Answer 1

 Need to choose 2 numbers a, b in such a way that   a+ b = -7     ab = 12* 6 =72

                                                                                               72=   36 . 2 = 18 . 4 = 9 . 8 = 3 . 24=12 .6

 

Factoring by completing the square:

 

 6 t^2 - 7t + 12 =

 

6 ( t ^2 - 7/6 t  +12/6) =

 

6[ ( t^2 - 2( 7/ 12) + 49/ 144 - 49/144 + 12/6)]

 

 6 [( t - 7/12) ^2  +239/144) =

 

  6 [( t- 7/12 + i(√239/12 ) ( t - 7/12 - i(√239/12 ) ]

 

Answer 2

6t^2 - 7t + 12 = (ax + b)(cx + d) = (ac)x^2 + (ad+bc)x +(bd)

ac = 6 = 1*2*3 = 1*6 = 2*3,
bd = 12 = 1*2*2*3 = 1*12 = 2*6 = 3*4,
ad+bc = -7

a . . . c . . . b . . . d . . . ad+bc
6 . . . 1 . . . -1 . . -12 . -72 -1 = -73
6 . . . 1 . . . -12 . . -1 . -6 -12 = -18
6 . . . 1 . . . -2 . . -6 . . -36 -2 = -38
6 . . . 1 . . . -6 . . -2 . . -12 -6 = -18
6 . . . 1 . . . -3 . . -4 . . -24 -3 = -27
6 . . . 1 . . . -4 . . -3 . . -18 -4 = -22
2 . . . 3 . . . -1 . . -12 . -24 -3 = -27
2 . . . 3 . . . -12 . . -1 . -2 -36 = -38
2 . . . 3 . . . -2 . . -6 . . -12 -6 = -18
2 . . . 3 . . . -6 . . -2 . . -4 -18 = -22
2 . . . 3 . . . -3 . . -4 . . -8 -9 = -17
2 . . . 3 . . . -4 . . -3 . . -6 -12 = -18
Having exhausted all the possible integer factors we can say the equation cannot be factored over the integers.

Let's use Vertex Form of:

f(t) = 6t^2 - 7t + 12 = 6(t - h)^2 + k

h = - b /(2a) = - -7/(2*6) = 7/12
k = f(h) = f(7/12) = ((6)(7/12) - 7)(7/12) + 12
= (42/12 - 84/12)(7/12) + 12
= (-42/12)(7/12) + 12
= -294/144 + 1728/144 = 1434/144 = 717/72 = 239/24

f(t) = 6(t - 7/12)^2 + 239/24 = 0
6(t - 7/12)^2 = -239/24
(t - 7/12)^2 = -239/144
t - 7/12 = +- i*sqrt(239)/12
t = 7/12 +- i*sqrt(239)/12

So in factored form:

f(t) = (t-(7/12 + i*sqrt(239)/12))(t-(7/12 - i*sqrt(239)/12))

Answer 3

The equation can not be factorized.

 

The only ways to solve it by the method of completing the square