(3x - 4)(2x^2 + 6x + 5)
6(ab - ac + b2 - bc)
6a to the second power minus 8ab + 2a
(x - 5)(x - 3)
6ab(2x2+x-5)
Remember to factor out the GCF of the coefficients if there is one. A perfect square binomial will always follow the pattern a squared plus or minus 2ab plus b squared. If it's plus 2ab, that factors to (a + b)(a + b) If it's minus 2ab, that factors to (a - b)(a - b)
(3y - 5)(y + 5)
(b-c)(a+b)-ac
(a - 2)(a^2 + 6)
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.
6(b - ac + b2 - bc)
6(ab - ac + b2 - bc)
No.
(x2 plus 40) (x minus 1)
That doesn't factor neatly. Applying the quadratic equation, we find two real solutions: (-2 plus or minus the square root of 5) divided by 3x = 0.07868932583326327x = -1.4120226591665965