This does readily factor!!!
So by trial and error.
Discount '0' because the you are left with '-7'
Try '-1
(-1)^(3) - 5(-1)^(2) - 13(-1) - 7 =>
-1 - 5 + 13 - 7 adding = '0'
Hence x = -1
Therefore ( x + 1) is a factor.
(x+1)
( x^(3) - 5(x^(2) - 13x - 7 = (x^(2)
-x^(3) -x^(2) =
-6x^(2) - 13x = ( x^(2) -6x
6x^(2) + 6x =
-7(x + 1) ( x^(2) - 6x + 7)
(x + 1)(x^(2) - 6x + 7) is fully factored.
If you factor the Quadratic, you will move into the region of IMGINARY (i) numbers.
5x^2 - 13x = x(5x - 13)
5x2-8x-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)
x(x - 13)(x - 1)
x3 + 13x2 + 42x = x(x2 + 13x + 42) = x(x2 + 6x + 7x + 42) = x[x(x + 6) + 7(x + 6)] = x(x + 7)(x + 6)
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7
5x^2 - 13x = x(5x - 13)
x2 • (5x2 + x + 8)
5x2-8x-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 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
x(x - 13)(x - 1)
x(x - 1)(x - 13)
x(x - 13)(x - 1)
x3 + 13x2 + 42x = x(x2 + 13x + 42) = x(x2 + 6x + 7x + 42) = x[x(x + 6) + 7(x + 6)] = x(x + 7)(x + 6)
2x2 - 13x - 7 = (2x + 1)(x - 7)
5x2 + 3x + 8x2 = 13x2 + 3x = x(13x + 3)