The question is equivalent to asking if there are two numbers whose sum is 6 (the coefficient of x) and whose product is 18 (the constant term). Since there are no such real numbers, the answer is NO. There is a solution in the complex domain but the normal context for such questions is the real numbers.
(x^18 - 6x^9 + 18)(x^18 + 6x^9 + 18)
xx+3x+7=6x+18 xx+3x+7-(6x+18)=6x+18-(6x+18) xx-3x-11=0 Factors of -11: 1,-11 -1,11 Doesn't factor evenly, use quadratic
I'm not sure this was notated correctly. As written, this is -11x plus or minus 18, which doesn't factor.
6x+4 factor: 2(3x+2)
The GCF is 6x.
(6x + 5)(6x + 5) or (6x + 5)2
(6x + 1)(x + 1)
6x(3x2 - x + 4)
Not factorable
(6x + 1)(x + 2)
(6x + 1)(x + 13)
x2 + 6x + 8 = (x + 4)(x + 2)