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
Done
According to the rational root theorem, the possible rational roots of the polynomial x^3 + 8x^2 + 6x - 6 = 0 can be found by considering the factors of the constant term (in this case, -6) divided by the factors of the leading coefficient (in this case, 1). So, the possible rational roots are:
±1, ±2, ±3, ±6
Among these, 4, 2, 5, and 1 are possibilities. Therefore, the answer is none of the given options.
To divide 3x^4 + 2x^3 - 5x^2 - 4 by x + 2, follow the long division method:
3x^3 - 4x^2 + 3x - 6
--------------------------------
x + 2 | 3x^4 + 2x^3 - 5x^2 - 4
- (3x^4 + 6x^3)
-------------
-4x^3 - 5x^2
+ ( -4x^3 - 8x^2 )
---------------------
3x^2 - 4x^2 + 3x - 4
+ (3x^2 + 6x)
---------------------
3x - 6 + 8/(x + 2)
Therefore, the division result is 3x^3 - 4x^2 + 3x - 6, with a remainder of 8/(x + 2).
To factor x^3 + 2x^2 - 9x - 18 given that -2 is a zero, we can use synthetic division:
-2 | 1 2 -9 -18
---------------------
1 0 -9 0
The result of synthetic division indicates that (x + 2) is a zero of the polynomial and the quotient is x^2 - 9.
Therefore, the factorization of x^3 + 2x^2 - 9x - 18 is (x + 2)(x^2 - 9) or (x + 2)(x - 3)(x + 3).
To determine which values are not possible rational roots of the polynomial x^3 + 8x^2 + 6, we can use the Rational Root Theorem.
The Rational Root Theorem states that if a polynomial has a rational root of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient, then p must be a factor of the constant term and q must be a factor of the leading coefficient.
For the polynomial x^3 + 8x^2 + 6 = 0, the constant term is 6 and the leading coefficient is 1. To find the factors of 6, we can list all possible combinations: ±1, ±2, ±3, ±6. To find the factors of 1, we have ±1.
Now, we can check each of these possible factors to see if they are indeed roots of the polynomial:
1: Substitute x = 1 into the polynomial: (1)^3 + 8(1)^2 + 6 = 1 + 8 + 6 = 15. Since it is not equal to zero, 1 is not a root.
2: Substitute x = 2 into the polynomial: (2)^3 + 8(2)^2 + 6 = 8 + 32 + 6 = 46. Again, it is not equal to zero, so 2 is not a root.
4: Substitute x = 4 into the polynomial: (4)^3 + 8(4)^2 + 6 = 64 + 128 + 6 = 198. Once again, it is not equal to zero, so 4 is not a root.
5: Substitute x = 5 into the polynomial: (5)^3 + 8(5)^2 + 6 = 125 + 200 + 6 = 331. It is not equal to zero, so 5 is not a root.
Therefore, out of the given options, the possible rational roots of the equation x^3 + 8x^2 + 6 = 0 are: 2, -1. The values 4, 5, and 1 are not possible rational roots.
For the second question, to divide the polynomial 3x^4 + 2x^3 - 5x^2 - 4 by x + 2, we can use long division:
___________________
x + 2 | 3x^4 + 2x^3 - 5x^2 - 4
- (3x^4 + 6x^3)
________________
- 4x^3 - 5x^2
+ ( - 4x^3 - 8x^2)
_________________
3x^2 - 4x -4
- (3x^2 + 6x)
______________
- 10x - 4
The result of the division is 3x^3 - 4x^2 + 3x - 6 with a remainder of -10x - 4, which can be written as (3x^3 - 4x^2 + 3x - 6) - (10x + 4) / (x + 2).
So the correct answer is 3x^3 - 4x^2 + 3x - 6 - (10x + 4) / (x + 2).
For the third question, to factor x^3 + 2x^2 - 9x - 18 given that -2 is a zero, we can use synthetic division.
Performing synthetic division with -2 as the zero, we have:
-2 | 1 2 -9 -18
-2 0 18
_________________________
1 0 -9 0
The result from the synthetic division is 1x^2 - 9, which can be factored as (x + 3)(x - 3).
Therefore, the correct factorization of x^3 + 2x^2 - 9x - 18 given that -2 is a zero is (x + 3)(x - 3)(x + 2).