Despite my serious qualms with the wording of your question, I am assuming it to be the factoring of x2-2x+3
This is not factorable. As long as there is a positive 3 in the position of the constant element, there can only be -4 or 4 for the coefficient of the single x. Perhaps you made an error in entering the question, so I will provide some likely alternatives.
If the trinomial read as x2-2x-3, then it would be factorable and would be (x-3)(x+1)
If the trinomial read as x2-4x+3, then it would be factored as (x-3)(x-1)
If the trinomial read as x2+4x+3, then it would be factored as (x+3)(x+1)
(2x+3)(x+3)
2(x + 3)(x + 1)
x2 + 2x - 15 = 0(x + 5) (x - 3) = 0x = -5x = +3
This is how I did it:8x3+4x2+6x+3[8x3+4x2]+[6x+3][(4x2)(2x+1)]+[(3)(2x+1)]....as you can see, this is in the general form of ab+cb, which factored is b(a+c), so:(2x+1)(4x2+3)
It is impossible but if it were x squared plus 2x minus 15 the equation would be (x+5) (x-3) with x being equal to either -5 or 3. If the original problem was x squared minus 2x minus 15 the equation would be (x-5)(x+3) and x would be equal to either 5 or -3
(x - 3)(2x - 7)
17
2x + 6 = 2(x+3)
x2-2x+3 does not factor over the field of real numbers. Over the complex numbers, it factors as (x - 1 +is)(x -1 -is) where s is the square root of 2.
It is 2(2x+3) when factorised
(5x + 2)(x + 3)
(2x+3)(x+3)
2x(y - 3) + 7(y - 3) = (2x + 7)(y - 3)
You cannot since the expression does not have rational roots.
(2x+3)(x+3)
(2x - 3)(2x - 3)
(2x+3)(2x+4)