2x(2x - 1)
(2x - 1)(4x + 1)
4(x - 1)(x + 1)
(4x - t2)= [(2√x)2 - t2] (This is now the difference of squares)= (2√x - t)(2√x + t)
4x2+4x-8 4(x2+x-2) and (4x+8)(x-1)
2x(2x - 1)
(2x - 1)(4x + 1)
(2x-1)(2x-1) = 4x^2 -4x + 1
(2x - 1)(2x + 1)
There is no rational factorisation.
4(x - 1)(x + 1)
(4x - 3)(2x + 3)
(2x - 4)(2x + 3) The key is to look for all the factor pairs of 12.
7x2 -4x -5x2 +2x = 2x2 -2x
(4x - t2)= [(2√x)2 - t2] (This is now the difference of squares)= (2√x - t)(2√x + t)
4x2+4x-8 4(x2+x-2) and (4x+8)(x-1)
8x2-10x-3 = (2x-3)(4x+1) when factored