Answer:
Fourth degree polynomial (aka: quartic)
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Work Shown:
There isnt much work to show here because we can use the fundamental theorem of algebra. The fundamental theorem of algebra states that the number of roots is directly equal to the degree. So if we have 4 roots, then the degree is 4. This is assuming that there are no complex or imaginary roots.
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If you want to show more work, then you would effectively expand out the polynomial
(x-m)(x-n)(x-p)(x-q)
where
m = 4, n = 2, p = sqrt(2), q = -sqrt(2)
are the four roots in question
(x-m)(x-n)(x-p)(x-q)
(x-4)(x-2)(x-sqrt(2))(x-(-sqrt(2)))
(x-4)(x-2)(x-sqrt(2))(x+sqrt(2))
(x^2-6x+8)(x^2 - 2)
(x^2-2)(x^2-6x+8)
x^2(x^2-6x+8) - 2(x^2-6x+8)
x^4-6x^3+8x^2 - 2x^2 + 12x - 16
x^4 - 6x^3 + 6x^2 + 12x - 16
We end up with a 4th degree polynomial since the largest exponent is 4.
Answer:
36
Step-by-step explanation:
You have to FOIL out the (x+4)(x-4) and then subrtract away the 9 in order to get a quadratic that you can solve for x. As it is, you can't do it.

Now if you move the 9 over with it, you get this:

which simplifies to

Now you can either solve this by recognizing that is the difference of perfect squares, or you can move the 25 over to the other side and take the square root of both sides, like this:


Answer:
B OR D
Step-by-step explanation: