To determine which of the given options is a factor of the polynomial \( x^4 + 6x^3 - 29x^2 - 114x + 280 \) given that \( x - 4 \) is a factor, we can perform polynomial long division or use synthetic division with \( x - 4 \) to find the quotient.
Let's start with synthetic division using \( x = 4 \):
- Set up the coefficients of the polynomial: \( [1, 6, -29, -114, 280] \).
- Use \( 4 \) in synthetic division:
4 | 1 6 -29 -114 280
| 4 40 44 -280
--------------------------
1 10 11 -70 0
The quotient we obtain is \( x^3 + 10x^2 + 11x - 70 \).
Now we need to factor \( x^3 + 10x^2 + 11x - 70 \) further. To find a root, we can test the potential rational roots. We start with integer values:
Let's test \( x = 2 \):
\[ f(2) = 2^3 + 10(2^2) + 11(2) - 70 = 8 + 40 + 22 - 70 = 0. \]
Since \( x = 2 \) is a root, we divide \( x^3 + 10x^2 + 11x - 70 \) by \( x - 2 \):
Perform synthetic division again with \( x = 2 \):
2 | 1 10 11 -70
| 2 24 70
---------------------
1 12 35 0
This means \( x^3 + 10x^2 + 11x - 70 \) factors as \( (x - 2)(x^2 + 12x + 35) \).
Next, we can factor \( x^2 + 12x + 35 \):
\[ x^2 + 12x + 35 = (x + 5)(x + 7). \]
Thus, the full factorization of the polynomial \( x^4 + 6x^3 - 29x^2 - 114x + 280 \) is:
\[ (x - 4)(x - 2)(x + 5)(x + 7). \]
Given the options, the factors we found, specifically \( x - 2 \) and \( x + 5 \) match with the potential choices. However, since \( x - 2 \) is explicitly listed, it is the correct answer.
Thus, the factor of \( x^4 + 6x^3 - 29x^2 - 114x + 280 \) is:
x - 2.