Originally Posted by
unstopabl3 a)
12x^2+x-6
12(x^2+x/12)-6
12{(x+1/24)^2+(1/24)(-1/24)}-6
12{(x+1/24)^2-1/576}-6
[tex]{12(x+1/24)^2+12(-1/576)}-6
{12(x+1/24)^2-1/48}-6
{12(x+1/24)^2-1/48-6(48/48)}
{12(x+1/24)^2-289/48}
12((x+1/24)^2-289/576)
12((x+1/24)^2-(17/24)^2) <<<<<<<<< difference of squares!
12(x+1/24+17/24)(x+1/24-17/24))
12(x+3/4)(x-2/3) <<<<<<<<< now split the factor 12 into 3 * 4 and put them into the appropriate bracket
(4x+3)(3x-2)
b)
4+6x=-x^2 <<<<<<<<< according to the wording of the problem
x^2+6x+4 = 0
(x^2+6x+9-9)+4=0
(x+3)^2-5=0
(x+3)^2-(sqrt(5))^2=0 <<<<<<<<<<< difference of squares
Factor the difference.
Aproduct equals zero if one factor equals zero.
Solve each equation for x.
Thanks!