x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
That depends on whether or not 2x is a plus or a minus
(a) y = -3x + 1
(x2 + 1)(x - 3)
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
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)
It is 4 terms of an algebraic expression.
== == 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
Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
That depends on whether or not 2x is a plus or a minus
(X + 1)3 = X3 + 3X2 + 3X + 1