your equation is this... x3 + 5x2 = 6x get all the terms on the left side x3 + 5x2 - 6x = 0 now factor out an "x" x(x2 + 5x - 6) = 0 now factor the equation inside x(x + 6)(x - 1) from this you can find your x-values x= 0, -6, and 1 now that you have your x-values your equation is solved. just use algebra to get all of the terms on the one side. then factor out anything that is common in all of the terms. then factor the polynomial. then find what values for "x" that satisfy the equation.
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
(x - 2)(x - 2)(x - 1)
your equation is this... x3 + 5x2 = 6x get all the terms on the left side x3 + 5x2 - 6x = 0 now factor out an "x" x(x2 + 5x - 6) = 0 now factor the equation inside x(x + 6)(x - 1) from this you can find your x-values x= 0, -6, and 1 now that you have your x-values your equation is solved. just use algebra to get all of the terms on the one side. then factor out anything that is common in all of the terms. then factor the polynomial. then find what values for "x" that satisfy the equation.
(x3+5x2+6x)/x to get x(x2+5x+6) find numbers that multiply to 6 and add to five (2 and 3) your factors are x, x+2, and x+3
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
(3x + 5)(x2 - 2)
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7
x3 + 3x2 - 6x - 8 = (x - 2)(x2 + 5x + 4) = (x - 2)(x + 1)(x + 4)
x3 + 4x2 + 6x + 24 = (x2 + 6)(x + 4)
True.
To factorise x3 + 5x2 - 16x - 80 I note that -80 = 5 x -16 and 5 & -16 are the coefficients of x2 and x. Thus I have: x3 + 5x2 - 16x - 80 = (x + 5)(x2 - 16) and the second term is a difference of 2 squares, meaning I have: x3 + 5x2 - 16x - 80 = (x + 5)(x + 4)(x - 4)
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
(x - 6)(x2 + 6x + 36)
-x(x2 - 2)(x2 + 3)