x3 + 3x2 - 9x + 5 = 0 has roots of -5,1 and 1. CHECK : x3 + 3x2 - 9x + 5 = (x + 5)(x - 1)(x - 1)
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
== == Suppose f(x) = x3 + 3x2 - 2x + 7 divisor is x + 1 = x - (-1); so rem = f(-1) = 11
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)
Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
y2=x3+3x2
x3 + 3x2 - 9x + 5 = 0 has roots of -5,1 and 1. CHECK : x3 + 3x2 - 9x + 5 = (x + 5)(x - 1)(x - 1)
x3 - 3x2 + x - 3 = (x2 +1)( x - 3)
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
== == Suppose f(x) = x3 + 3x2 - 2x + 7 divisor is x + 1 = x - (-1); so rem = f(-1) = 11
x3 + 3x2 - 6x - 8 = (x - 2)(x2 + 5x + 4) = (x - 2)(x + 1)(x + 4)
(x3)'=3x2(x3)''=(3x2)'=6x
x3 /12 + 16x + c
Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
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)
That depends on whether or not 2x is a plus or a minus