if y = x^3 +x^2+x-3, finding the roots of y can help. the roots are when x^3 +x^2+x-3 = 0, which x = 1 satisfies. Using something called Horner's method of synthetic division, (look it up, it's not too bad), you can take out the (x-1) and have (x-1)(x^2+2x+3) = 0, so either x - 1 = 0, x =1 (from before), or (x^2+2x+3) = 0, which you can factorise using quadratic methods to get x = (-2 (+or-) sqrt(4 - 12))/2 = 1 +- sqrt(2)i, so its not a pretty answer. involves complex numbers, I think this is right ayway .
x³ - x² + x - 3 (x² + 1)(x - 1) - 2
(3 - i)(3 + i)
(a - 2)(a^2 + 6)
x(x + 3)(x - 4)
Without that other number, all you've got is zero.
Factor out the GCF and get X(X2-X+1).
To factor out the expression: x2y-y3 First factor out one "y": y(x2-y2) The expression x2-y2 is a difference of squares, which factors as well: (y)(x-y)(x+y) This is the simplest factoring of the original expression.
(2x3 - x2) = x2 (2x - 1)
(x + 6)(x - 6)
(x - 2)(x^2 + x + 3)
(x + 2)(3x - 1)(3x + 1)
(x - 5)(x^2 + 1)