Answer: Yes, (x+4) is a factor of the given polynomial.
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Explanation:
We can use the remainder theorem here.
Consider any polynomial p(x). The binomial x-k will be a factor of p(x) if and only if p(k) = 0. I'm using the special case of the remainder theorem, where we look for a remainder of 0.
The binomial x+4 is the same as x-(-4). Compare this to x-k to see that k = -4.
Then plug this k value into the p(x) polynomial given
p(x) = 2x^3 + 16x^2 + 22x - 40
p(-4) = 2(-4)^3 + 16(-4)^2 + 22(-4) - 40
p(-4) = 2(-64) + 16(16) + 22(-4) - 40
p(-4) = -128 + 256 - 88 - 40
p(-4) = 0
This shows that dividing p(x) = 2x^3 + 16x^2 + 22x - 40 over x+4 leads to a remainder of 0. Therefore, (x+4) is a factor of p(x).
