Answer: 3 x^4 - x^3 - 3 x^2 + 1 = (x - 1)(3 x^3 + 2 x^2 - x - 1)
Explanation:
Let's denote our polynomial 3 x^4 - x^3 - 3 x^2 + 1 by P(x).
We use the Remainder Theorem: when dividing a polynomial P(x) by (x-c), where c is a constant, the remainder equals P(c).
Hence, for x - c to be a divisor of the polynomial P(x), the remainder P(c) needs to be equal to zero.
Therefore, we need to find a constant c such that P(c) = 0.
P(c) = 0 <=> 3 c^4 - c^3 - 3 c^2 + 1 = 0 <=> 3 c^2 (c^2 - 1) - (c^3 - 1)=0 <=> 3 c^2(c+1)(c-1) - (c - 1)(c^2 + c + 1)=0 <=> (c - 1)(3 c^3 + 3 c^2 - c^2 - c - 1)=0 <=> (c - 1)(3 c^3 + 2 c^2 - c - 1)=0
So, we found that, if c=1, then P(c) = 0. Therefore, x -1 is a divisor of the polynomial P(x).
Verification : divide P(x)=3 x^4 - x^3 - 3 x^2 + 1 by x - 1 and obtain
P(x)/(x - 1)=3 x^3 + 2 x^2 - x - 1