It is: 4x(1+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)
4(x + 2)(x + 2)
x + 4
x3 + 4x2 - 16x - 64 = (x + 4)(x + 4)(x - 4).
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
x2+16x-80 = 0 (x-4)(x+20) = 0 x = 4 or x = -20
x(x + 4)(x - 4) or x3 - 16x
x2-16x+48 = (x-4)(x-12) when factored
It is: 4x(1+4)
They are 4 terms of an expression that can be simplified to: 4 -15x^2
x3 /12 + 16x + c
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
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)
4(x + 2)(x + 2)
Not factorable
x + 4