If it's (x3 -1) that you want to factorize, then find the solutions to (x3 -1) = 0.
So if P(x) is a polynomial of x, and x=a is a solution for P(x) = 0, then (x - a) is a factor of P(x).
So x = 1 solves (x3 -1) = 0, so (x - 1) is a factor. Use long division (x3 -1)/(x-1) = x2 + x + 1. Use the quadratic formula to find the roots of this: -1/2 ± i*sqrt(3)/2, which is complex. So the factorization is:
Multiply the polynomials together to check that your answer is correct.
It depends on what you wanted to do - graph it, solve it, factorise it, etc.
6
It could be x(1 -1) when factorised
x(x - 7)
( x - 14 ) ( x + 14 )
X3 + 8= (x+2)(x2-2x+4)
x3-16xx(x2-16)x(x+4)(x-4)...forgot to finish it, sorry.
6x^3 + 3x = 3x(2x^2 + 1)
It depends on what you wanted to do - graph it, solve it, factorise it, etc.
x3 - 9x = x(x2 - 9) = x(x2 - 32) = x(x - 3)(x + 3)
X cubed - X cubed is zero.
6
x + x = x(1+1)
xcubed-1 Answer::(X-1)(Xsquared+X+1) when you factor xcubed minus a number its the same thing as x cubed minus y cubed and x cubed minus y cubed factors to:: (x-y)(xsquared+xy+y squared) the first factor, (x-y), is the cubed root of the first and the cubed root of the second, so in the answer i have (x-1), which is x cubed minus one cubed :) the second factor, (xsquared+xy+ysquared), you take the first number squared, Xsquared, then the first and second one multiplied together, XY, and then the second number squared, Ysquared, so in the answer i have (xsquared+x+1), which is x squared, then x times 1 which is just x, and positive 1, which is negative 1 squared :) x^3 - 1
It is: x(x-3) factorized
It could be x(1 -1) when factorised
x(x - 7)