Question 1004098
I'll do the first problem to get you started.



The best way to do this problem is to check each answer choice


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Let's check choice A. This factor is x+2, so solve x+2 = 0 to get x = -2


Plug in x = -2 and evaluate


y = x^4-27x^2-14x+120


y = (-2)^4-27(-2)^2-14(-2)+120 ... replace each x with -2


y = 56


The result y = 56 is NOT zero, so x = -2 is NOT a root of the polynomial.


Since x = -2 is NOT a root of the polynomial, this means x+2 is NOT a factor of the polynomial.



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Let's check choice B. This factor is x-3, so solve x-3 = 0 to get x = 3


Plug in x = 3 and evaluate


y = x^4-27x^2-14x+120


y = (3)^4-27(3)^2-14(3)+120 ... replace each x with 3


y = -84


The result y = -84 is NOT zero, so x = 3 is NOT a root of the polynomial.


Since x = 3 is NOT a root of the polynomial, this means x-3 is NOT a factor of the polynomial.



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Let's check choice C. This factor is x+4, so solve x+4 = 0 to get x = -4


Plug in x = -4 and evaluate


y = x^4-27x^2-14x+120


y = (-4)^4-27(-4)^2-14(-4)+120 ... replace each x with -4


y = 0


The result y = 0 is zero, so x = -4 is definitely a root of the polynomial.


Since x = -4 is a root of the polynomial, this means x+4 is a factor of the polynomial.



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Let's check choice D. This factor is x+5, so solve x+5 = 0 to get x = -5


Plug in x = -5 and evaluate


y = x^4-27x^2-14x+120


y = (-5)^4-27(-5)^2-14(-5)+120 ... replace each x with -5


y = 140


The result y = 140 is NOT zero, so x = -5 is NOT a root of the polynomial.


Since x = -5 is NOT a root of the polynomial, this means x+5 is NOT a factor of the polynomial.



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Summary:


Only x+4 is a factor of the polynomial. So only choice C is the answer.