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The remainder when x^4 + 2x^3 + 3x^2 + 4x + 5 is divided by x + 2 is -3.

To find the remainder when a polynomial is divided by a linear factor, we can use the remainder theorem. This states that if we divide a polynomial f(x) by x - a, the remainder is f(a).

In this case, we want to divide x^4 + 2x^3 + 3x^2 + 4x + 5 by x + 2. Using long division, we get:

x^3 - 2x^2 + 7x - 14

x + 2 | x^4 + 2x^3 + 3x^2 + 4x + 5

- x^4 - 2x^3

---------------

x^2 + 4x

- x^2 - 2x

-----------

2x + 5

- 2x - 4

-------

9

Therefore, the remainder is 9. However, we need to use the remainder theorem to find the remainder when x = -2. Plugging in -2 for x in the original polynomial, we get:

(-2)^4 + 2(-2)^3 + 3(-2)^2 + 4(-2) + 5 = 16 - 16 + 12 - 8 + 5 = -3

Therefore, the remainder when x^4 + 2x^3 + 3x^2 + 4x + 5 is divided by x + 2 is -3.

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