x2+7x+12
Multiply the coefficient of x2 and the constant.
12 = 4 x 3 and when you add 4 and 3 you get 7 which is the co efficient of x.
Therefore the above eqn can be written as
x2 + 4x + 3x + 12
= x(x+4) + 3(x+4)
= (x+3)(x+4)
x3-x = x(x2-1)
= x(x+1)(x-1)
16 - 14t - 2t2
Multiply the coefficient of t2 and the constant
16 x 2 = 32
Represent 32 = 16 x 2 and when you subtract them (16-2) you get 14 which the co efficient of t.
16 - 16t + 2t - 2t2
16(1-t) + 2t(1-t)
(16+2t)(1-t)
2(8+t)(1-t)
x4 - 2x2 - 8
Multiply the coefficient of x4 and the constant
1 x 8 = 8
Represent 8 = 4 x 2 and when you subtract them (4-2) you get 2 which the co efficient of x2.
x4 - 4x2 + 2x2 - 8
x2(x2-4) + 2(x2-4)
(x2+2)(x2-4)
(x2+2)(x-2)(x+2)
(