How do you factor 3x^4-x^3-3x^2+13x4x33x2+1?

2 Answers
May 18, 2015

We can see clearly that all elements that contain the variable xx are multiplying x^2x2, so we can factor it out as follows:

x^2(3x^2-x-3)+1x2(3x2x3)+1

x^2[x(3x-1)-3]+1x2[x(3x1)3]+1

May 18, 2015

Answer: 3 x^4 - x^3 - 3 x^2 + 13x4x33x2+1 == (x - 1)(3 x^3 + 2 x^2 - x - 1)(x1)(3x3+2x2x1)

Explanation:

Let's denote our polynomial 3 x^4 - x^3 - 3 x^2 + 13x4x33x2+1 by P(x)P(x).

We use the Remainder Theorem: when dividing a polynomial P(x)P(x) by (x-c)(xc), where cc is a constant, the remainder equals P(c)P(c).

Hence, for x - cxc 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 3c4c33c2+1== 00 <=> 3 c^2 (c^2 - 1) - (c^3 - 1)3c2(c21)(c31)==00 <=> 3 c^2(c+1)(c-1) - (c - 1)(c^2 + c + 1)3c2(c+1)(c1)(c1)(c2+c+1)==00 <=> (c - 1)(3 c^3 + 3 c^2 - c^2 - c - 1)(c1)(3c3+3c2c2c1)==00 <=> (c - 1)(3 c^3 + 2 c^2 - c - 1)(c1)(3c3+2c2c1)==00

So, we found that, if cc==11, then P(c) = 0P(c)=0. Therefore, x -1x1 is a divisor of the polynomial P(x)P(x).

Verification : divide P(x)P(x)==3 x^4 - x^3 - 3 x^2 + 13x4x33x2+1 by x - 1x1 and obtain
P(x)P(x)//(x - 1)(x1)==3 x^3 + 2 x^2 - x - 13x3+2x2x1