Question

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

Answer 1

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

`x^2(3x^2-x-3)+1`

`x^2[x(3x-1)-3]+1`

Answer 2

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`