Question

3, if 2 is the root of p(x) =x^3-2x^2+4x+8. then find the remaining roots of p(x).

Answer 1

if 2 is a root, dividing by x-2 will yield a remainder of zero. Since it did not, x=2 is not a root.
So the problem is in error.

In fact, there are no rational roots, and two roots are complex.

To check your work, (x-2)(x-2)(x+2) = x^3 - 2x^2 - 4x + 8
so apparently the polynomial had a typo in it.

Answer 2

To find the remaining roots of the polynomial, we can use synthetic division or polynomial division to divide the given polynomial by (x-2) which is the factor corresponding to x=2 being a root.

Using synthetic division:

2 | 1 -2 4 8
| 2 0 8
-----------------
1 0 4 16

The quotient is x^2 + 4, which can be factored as (x+2)(x-2).

Therefore, the remaining roots of the polynomial are x=-2 and x=2.

Answer 3

I apologize for the mistake in the original problem statement. Thank you for pointing it out.

Given the corrected polynomial: p(x) = x^3 - 2x^2 - 4x + 8

Since there are no rational roots, we need to find the remaining roots using another method such as synthetic division or polynomial division to factor the polynomial further.

Let's perform synthetic division to factorize the polynomial:

We try divisions by potential factors of 2, by (x-2), we get:

2 | 1 -2 -4 8
| 2 0 -8
|_______________
1 0 -4 0

Therefore, the factored form of the polynomial is: p(x) = (x-2)(x^2 - 4)

Further factoring:
p(x) = (x-2)(x+2)(x-2)

Thus, the roots of the polynomial p(x) = x^3 - 2x^2 - 4x + 8 are x=2 (double root) and x=-2.