Answer:
\((x-5)\)
Step-by-step explanation:
The Factor Theorem tells us that substituting \(x=a\) into a polynomial \(p(x)\) when divided by a linear factor \((x-a)\) will give us a value of 0 IF \((x-a)\) is a factor of the polynomial \(p(x)\)
So, checking each of the 4 binomials, and getting a 0 as answer will tell us which one is a factor of the function given.
Putting \(x=-1\) for \((x+1)\) into the function:
\(3(-1)^{3}-12(-1)^2-4(-1)-55\\=-66\)
NOT a factor.
Putting \(x=-4\) for \((x+4)\) into the function:
\(3(-4)^{3}-12(-4)^2-4(-4)-55\\=-423\)
NOT a factor.
Putting \(x=5\) for \((x-5)\) into the function:
\(3(5)^{3}-12(5)^2-4(5)-55\\=0\)
IS a factor.
Putting \(x=2\) for \((x-2)\) into the function:
\(3(2)^{3}-12(2)^2-4(2)-55\\=-87\)
NOT a factor.
So we can see that only \((x-5)\) is a factor.
