If some linear binomial x-a is a factor of polynomial [tex] f(x) = x^3 + 6x^2 + 12x + 35 [/tex], then f(a)=0.
Let check it:
1. For the binomial x-1, you have that a=1 and
[tex] f(1) = 1^3 + 6\cdot 1^2 + 12\cdot 1 + 35=54\neq 0. [/tex]
2. For the binomial x-3, you have that a=3 and
[tex] f(3) = 3^3 + 6\cdot 3^2 + 12\cdot 3 + 35=27+54+36+35=152\neq 0. [/tex]
3. For the binomial x+3, you have that a=-3 and
[tex] f(1) = (-3)^3 + 6\cdot (-3)^2 + 12\cdot (-3)+ 35=-27+54-36+35=26\neq 0. [/tex]
4. For the binomial x+5, you have that a=-5 and
[tex] f(1) = (-5)^3 + 6\cdot (-5)^2 + 12\cdot (-5) + 35=-125+150-60+35=0. [/tex]
Answer: correct choice is x+5.
