8x2-3x-5 = (8x+5)(x-1) when factored
3(8x + 1)(8x - 1)
5 - 8x + 3x2 = 3x2 - 8x + 5 (since the sign of the second term is negative, then 5 can be factored as (-1)(-5)) = (3x - )(x - ) (try to put 1 or 5 to the empty places, in order to obtain -8x) = (3x - 5)(x - 1)
(4x+1)(4x-3)
4x^(2) - 8x '4' is common to both '4' & '8' . So tak outr '4'. Hence 4(x^(2) - 2x) Inside the brackets 'x' is common to both 'x^(2) & '2x'. So take out 'x'. Hence 4x(x - 2) Done , fully factored.
(2x - 1)(4x + 1)
(8x + 9)(3x^2 - 1)
(2x + 1)(4x^2 - 2x + 1)
(x-3)(3x+1)
t3-1/27
(x - 1)(x - 1)(x - 1) - (x + 1)(x + 1)(x + 1)
x(x + 1)(x -1)
(x - 1)(x^2 + x + 1)
8(y - 1)(y^2 + y + 1)
(x - 1)(x^2 + x + 1)
8x2-3x-5 = (8x+5)(x-1) when factored
(x + 2)(3x - 1)(3x + 1)