This is a quadratic equation question to find the value of x. x2-x-6 = 0 (x+2)(x-3) = 0 So x = -2 or +3
(x - 2)(x - 3)
x2 + 7x + 6 = (x + 6) (x + 1)
x2 + 5x - 6 can be rewritten as (x+1)(x-6). Setting each bracket = 0 gives x = -1 or x=6.
6
x2 + x = 6, therefore x2 + x - 6 = 0This can be factored : x2 + x - 6 = (x + 3)(x - 2) = 0This equation holds true when x + 3 = 0 or x - 2 = 0.When x + 3 = 0 then x = -3 : When x - 2 = 0 then x = 2
x2 - 5x - 6 = x2 - 6x + x - 6 = x(x - 6) + 1(x - 6) = (x - 6)(x + 1)
x2 + 5x - 6 = x2 + 6x - x - 6 = x(x + 6) - 1(x + 6) = (x - 1)(x + 6)
-x2 = x - 6 x2 + x - 6 = 0 (x - 2)(x + 3) = 0 x ∈ {-3, 2}
x2 - 36 = x2 - 62 = (x + 6)(x - 6)
x2 + x - 6 = (x + 3)(x - 2)
-4
-x2 + x + 42 = -[x2 - x - 42] = -[x2 - 7x + 6x - 42] = -[x*(x - 7) + 6*(x - 7)] = -(x - 7)*(x + 6)
x2+x-6=x2+3x-2x-6x2+x-6=x(x+3)-2(x-3)x2+x-6=(x+3)(x-2)
x2 + 3x - 18 = x2 + 6x - 3x - 18 = x(x + 6) - 3(x + 6) = (x + 6)(x - 3)
(x - 2)(x - 3)
x2+x-6 = (x-2)(x+3) when factored
38