How do you factor the trinomial -x^5 + 2x^3 + x - 1−x5+2x3+x−1?

1 Answer
George C.
Nov 2, 2016

-x^5+2x^3+x-1 = -(x^2+x-1)(x^3-x^2-1)−x5+2x3+x−1=−(x2+x−1)(x3−x2−1)

Explanation:

Given:

-x^5+2x^3+x-1−x5+2x3+x−1

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

-x^5+2x^3+x-1 = -(x^5-2x^3-x+1)−x5+2x3+x−1=−(x5−2x3−x+1)

Let:

f(x) = x^5-2x^3-x+1f(x)=x5−2x3−x+1

By the rational roots theorem, any rational zeros of f(x)f(x) are expressible in the form p/qpq for integers p, qp,q with pp a divisor of the constant term 11 and qq a divisor of the coefficient 11 of the leading term.

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

We find:

f(1) = 1-2-1+1 = -1f(1)=1−2−1+1=−1

f(-1) = -1+2+1+1 = 3f(−1)=−1+2+1+1=3

So f(x)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 11 and the constant term 11, there are two possible forms to consider:

(1) " "x^5-2x^3-x+1 = (x^2+ax+1)(x^3+bx^2+cx+1) x5−2x3−x+1=(x2+ax+1)(x3+bx2+cx+1)

(2) " "x^5-2x^3-x+1 = (x^2+ax-1)(x^3+bx^2+cx-1) x5−2x3−x+1=(x2+ax−1)(x3+bx2+cx−1)

Here's what we get when we look at (2), long dividing x^5-2x^3-x+1x5−2x3−x+1 by (x^2+ax-1)(x2+ax−1)...

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Notice the remainder is:

(a^4+a^2-2)x+(1-a^3)(a4+a2−2)x+(1−a3)

We can make this 00 by setting a=1a=1, finding that:

x^5-2x^3-x+1 = (x^2+x-1)(x^3-x^2-1)x5−2x3−x+1=(x2+x−1)(x3−x2−1)

Then negating both sides to solve the original problem:

-x^5+2x^3+x-1 = -(x^2+x-1)(x^3-x^2-1)−x5+2x3+x−1=−(x2+x−1)(x3−x2−1)

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