x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
1
(a + 6)(a + 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)
p2+10d+7
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
X3 + 8= (x+2)(x2-2x+4)
-24
X(X2 - X)
1
(a + 6)(a + 4)
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7
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=5
p2+10d+7
Anything plus 0 is itself, so x3 + 0 is x3.
x3-16xx(x2-16)x(x+4)(x-4)...forgot to finish it, sorry.