Given 4x2+4x-1
Using the formula for the roots of quadratic equation, (-b +/-./b2-4ac)/2a
the roots for the above will be (-4+/-4./2)/8 = (-1+/-./2)/2
Hence the two roots are (-1+./2)/2 and (-1-./2)/2
As the roots are irrational, there is no possibility of getting factors.
(2x-1)(2x-1) = 4x^2 -4x + 1
It is 9*(4x + 1).
16x2 + 8x + 1 - 9y2 = (4x + 1)2 - (3y)2 which is a difference of two squares. = (4x + 1 + 3y)*(4x + 1 - 3y)
3(x2 + 3x + 1)
(x + 1)(2x + 5)
(2x-1)(2x-1) = 4x^2 -4x + 1
4x(4x^2 + 3x + 1)
2(x - 1)(2x + y)
There is no rational factorisation.
(4x + 1)(2x + 3)
(2x - 1)(6x + 5)
x2 + 4x + 3 =(x + 3) (x + 1)
The answer is (2x^2+3)(4x+1)
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
4x2+4x-8 4(x2+x-2) and (4x+8)(x-1)
x = -½
B squared -4ac is the discriminant. you get 0, which means it will have one rational solution.