6x2 + 11x - 10 = (6x2 + 15x) - (4x + 10)
= 3x(2x + 5) - 2(2x + 5)
= (3x - 2)(2x + 5).
Note, first, that the three co-efficients of the original quadratic expression are 6, 11, and -10.
We need two numbers whose sum is 11 and
whose product is (6)(-10) = -60.
These two numbers turn out to be 15 and -4.
Thus, we replace the middle term, 11x with 15x - 4x and proceed with the factorisation in the usual way.
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-3x3 - 6x2 + 189x = -3x(x2 + 2x - 63) = -3x(x + 9)(x - 7)
6x2 + 11x + 3 = 6x2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1)
6x2 + 10x = 2x*(3x + 5)
6x2-5x-25 = (3x+5)(2x-5) when factored
This doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: (-17 plus or minus the square root of -671) divided by 12x = -1.4166 + 2.158638974498103ix = -1.4166 - 2.158638974498103iwhere i is the square root of negative one.