a(4a - 3)(3a - 5)
(3y - 5)(y + 5)
(b-c)(a+b)-ac
(a - 2)(a^2 + 6)
3x^(2) +9x - 2x -6 Collect 'like terms'. Hence 3x^(2) + 7x - 6 Next write down all the factors of '3' and '6' Hence 3 ; 1' & 3' 6 ; 1,6 ; 2,3. From these pairs of number we select a pair from each coefficient, that add/multiply to '7' . Hence (3' x 3 ) & (1' x 2) ; NB 'dashes' (') to indicate source of numbers. Write up brackets (3x 2)(x 3) -2)(x + 3) Next we notice that the '6' is negative, so the two signs are different (+/-) The '7x' is positive , so the larger number takes the positive sign . Hence (3x - 2)(x + 3)
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)
No.