Answer: 3 x^4 - x^3 - 3 x^2 + 13x4−x3−3x2+1 == (x - 1)(3 x^3 + 2 x^2 - x - 1)(x−1)(3x3+2x2−x−1)
Explanation:
Let's denote our polynomial 3 x^4 - x^3 - 3 x^2 + 13x4−x3−3x2+1 by P(x)P(x).
We use the Remainder Theorem: when dividing a polynomial P(x)P(x) by (x-c)(x−c), where cc is a constant, the remainder equals P(c)P(c).
Hence, for x - cx−c to be a divisor of the polynomial P(x)P(x), the remainder P(c)P(c) needs to be equal to zero.
Therefore, we need to find a constant cc such that P(c) = 0P(c)=0.
P(c) = 0P(c)=0 <=>⇔ 3 c^4 - c^3 - 3 c^2 + 1 3c4−c3−3c2+1== 00 <=>⇔ 3 c^2 (c^2 - 1) - (c^3 - 1)3c2(c2−1)−(c3−1)==00 <=>⇔ 3 c^2(c+1)(c-1) - (c - 1)(c^2 + c + 1)3c2(c+1)(c−1)−(c−1)(c2+c+1)==00 <=>⇔ (c - 1)(3 c^3 + 3 c^2 - c^2 - c - 1)(c−1)(3c3+3c2−c2−c−1)==00 <=>⇔ (c - 1)(3 c^3 + 2 c^2 - c - 1)(c−1)(3c3+2c2−c−1)==00
So, we found that, if cc==11, then P(c) = 0P(c)=0. Therefore, x -1x−1 is a divisor of the polynomial P(x)P(x).
Verification : divide P(x)P(x)==3 x^4 - x^3 - 3 x^2 + 13x4−x3−3x2+1 by x - 1x−1 and obtain
P(x)P(x)//(x - 1)(x−1)==3 x^3 + 2 x^2 - x - 13x3+2x2−x−1
