Question

How do you factor the trinomial `-x^5 + 2x^3 + x - 1`?

Answer 1

`-x^5+2x^3+x-1 = -(x^2+x-1)(x^3-x^2-1)`

Explanation:

Given:

`-x^5+2x^3+x-1`

First separate out the scalar factor `-1` to give us a positive leading coefficient:

`-x^5+2x^3+x-1 = -(x^5-2x^3-x+1)`

Let:

`f(x) = x^5-2x^3-x+1`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `1` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are `+-1`

We find:

`f(1) = 1-2-1+1 = -1`

`f(-1) = -1+2+1+1 = 3`

So `f(x)` has no rational zeros and no linear factors with rational coefficients.

So if we want rational coefficients, we are looking for the product of a quadratic and a cubic.

Further, since the leading coefficient is `1` and the constant term `1`, there are two possible forms to consider:

(1) `" "x^5-2x^3-x+1 = (x^2+ax+1)(x^3+bx^2+cx+1)`

(2) `" "x^5-2x^3-x+1 = (x^2+ax-1)(x^3+bx^2+cx-1)`

Here's what we get when we look at (2), long dividing `x^5-2x^3-x+1` by `(x^2+ax-1)`...

Notice the remainder is:

`(a^4+a^2-2)x+(1-a^3)`

We can make this `0` by setting `a=1`, finding that:

`x^5-2x^3-x+1 = (x^2+x-1)(x^3-x^2-1)`

Then negating both sides to solve the original problem:

`-x^5+2x^3+x-1 = -(x^2+x-1)(x^3-x^2-1)`