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

1 Answer
Nov 2, 2016

-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)...

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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)