(11x - 2)(3x - 5)
(x + 3)(3x - 2)
(3y - 5)(y + 5)
No.
(x2 plus 40) (x minus 1)
(x + 7)(x - 6)
(x-4)
(3y - 5)(y + 5)
(b-c)(a+b)-ac
(a - 2)(a^2 + 6)
(x + 3)(3x - 2)
That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: 1 plus or minus i where i is the square root of negative one.
2(a + 7)(a - 6)
There are no rational factors.
There is no rational factorisation.
(3x - 4)(2x^2 + 6x + 5)
6(b - ac + b2 - bc)
6(ab - ac + b2 - bc)