Anything plus 0 is itself, so x3 + 0 is x3.
x(x + 2)(x + 3) x = 0, -2, -3
(x - 2)(x + 3)(x + 5)
(x + y)(x + y)(x + y)
x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)
You have to put your heart into it!
x^3 - x^3 = 0 Remember , whilst 'x' is an unknown value, that unkonwn is a fixed value. As a numerical example 3^(3) - 3^(3) = 27 - 27 = 0 The '3' is 'x' in this case
no
x3 + x = x2(x + 1)
x(x^2 + 1)
x3+8y3 = (x+2y)(x2-2xy+4y2) The discriminant of the quadratic factor is 4y2-16y2 < 0 so there are no real roots. So the only real root of the original polynomial is x+2y=0 or x = -2y
x + x = 2x. Similarly, x^3 + x^3 = 2(x^3)
x(x2 + 36)