x3 + 8 = x3 + 23 = (x + 2)[x2 - (x)(2) + 22] = (x + 2) (x2 - 2x + 4)
x2(x3 + 1) is the best you can do there.
x3 + 1 = x3 + x2 - x2 - x + x + 1 = x2(x + 1) - x(x + 1) +1(x + 1) = (x + 1)(x2 - x + 1)
1.6667
x3 - x2 - 2x +2 you can divide this into two parts for now x3 - x2 and -2x +2 factor out everything you can from both x2(x-1) and -2(x+1) since the insides of the parentheses are the same, we can put them together, then multiply it by the sum of the two parts outside the parentheses. (x-1)(x2-2)
If x2 and x3 are meant to represent x2 and x3, then x2 times x3 = x5 You find the product of exponent variables by adding the exponents.
x3-x2
X3 X(X2) X2(X) and, X * X * X
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
x3 - x2 + 2x = x*(x2 - x + 2) which cannot be factored further.
What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
x3 - x2 + x - 2 has no rational factors.
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
Answer: x (x2-x)
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)