Question

According to the rational root theorem, which is not a possible rational root of x3 + 8x2 x 6 = 0?

4
2 <-- or 1?
5
1

What os 3x^4+2x^3-5x^2-4 divided by x+2

3x^3-4x^2+3x-6+8/x+2 <--?
3x^3-4x^2+3x-6+8/x-2

factor x^3+2x^2-9x-18 given that -2 is a zero

(x+2)(x-sqrt3)(x+sqrt3)=0
(x-2)(x-3)(x+3)=0
(x+2)(x-3)(x+3)=0 <--?
(x-2)(x^2-9)=0

Answer 1

Please number your question, and use proper notation for powers

1.
x^3 + 8x^2 + 6 = 0 , (I assumed the + sign at the end)
Without any work, if we assume it does factor, then any factor CANNOT end in a number which is not a factor of 6, since the 6 can only arrive from the multiplication of the end terms of your factors.
6 = (1)(2)(3), so it cannot possible end in a 5 or 4

2.
(3x^4+2x^3-5x^2-4)/(x+2) <--- those brackets are essential!!
(x+2) does not divide evenly,
I got
(3x^4+2x^3-5x^2-4)/(x+2) = (3x^3 - 4x^2 + 3x - 6) + 8/(x+2)
other way to show it does not divide evenly:
f(-2) = 3(16) + 2(-8) - 5(4) - 4
= 8 , which was the remainder I had above.

3. x^3+2x^2-9x-18 , if -2 is a zero ---> x+2 is a factor.
We could just factor it by grouping
x^3+2x^2-9x-18
= x^2(x + 2) - 9(x+2)
= (x^2 - 9)(x+2)
= (x+3)(x-3)(x+2)