X^2 - 13X + 30 = 0 (X-3)(X-10) easy inspection problem
It's a quadratic expression: x²+13x+12 If, for example, x²+13x+12 = 0 then there are 2 roots, x=-1 and x=-12
2x2 - 13x - 7 = (2x + 1)(x - 7)
x2 + 13x + 36 = 0 so (x+4)(x+9) = 0 so that x = -4 or x = -9
2x - 13x + 42 = x +ax + b a + b = 2(x - 6.5x + 21) = 34 = a + b
(x - 5)(x - 8)
(x + 5)(x + 8)
x2 + 13x + 36 = (x + 9)(x + 4)
-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)
2x2+13x+15 = (2x+3)(x+5)
(x + 4)(x + 9)
(x + 1)(3x + 10)
You should look for two numbers whose product is 40, and whose sum is 13. Experimenting a little (trying different pairs of factors for 40), you can quickly realize that this works for 8 and 5. Therefore, x2 + 13x + 40 = (x + 8) (x + 5)
(x + 9)(x + 4)
(x + 10)(x + 3)
x(x - 13)(x - 1)
It is x^2 -13x +12 = (x-1)(x-12) when factored