Which binomial is a factor of f(x) = x^4 27x^2 14x + 120?
(x 4)
(x 2)
(x + 5)
(x + 6)
you have a problem with - signs?
x^4 - 27x^2 - 14x + 120
= (x+4)(x+3)(x-2)(x-5)
so, I guess (B)
To find out which binomial is a factor of the given polynomial, we can use the Remainder Theorem. According to the Remainder Theorem, if a binomial (x - a) is a factor of a polynomial f(x), then f(a) = 0.
Let's test each binomial option by substituting it in for x in the polynomial and checking if the result is 0.
1. For the binomial (x - 4):
f(4) = (4)^4 + 27(4)^2 + 14(4) + 120 = 256 + 432 + 56 + 120 = 864 + 176 = 1040. Since f(4) is not equal to 0, (x - 4) is not a factor.
2. For the binomial (x - 2):
f(2) = (2)^4 + 27(2)^2 + 14(2) + 120 = 16 + 108 + 28 + 120 = 124 + 148 = 272. Since f(2) is not equal to 0, (x - 2) is not a factor.
3. For the binomial (x + 5):
f(-5) = (-5)^4 + 27(-5)^2 + 14(-5) + 120 = 625 + 675 - 70 + 120 = 1300. Since f(-5) is not equal to 0, (x + 5) is not a factor.
4. For the binomial (x + 6):
f(-6) = (-6)^4 + 27(-6)^2 + 14(-6) + 120 = 1296 + 972 - 84 + 120 = 2292 - 84 + 120 = 2336. Since f(-6) is not equal to 0, (x + 6) is not a factor.
Therefore, none of the given binomials is a factor of f(x) = x^4 + 27x^2 + 14x + 120.