How do you factor the trinomial -x^5 + 2x^3 + x - 1?
1 Answer
Explanation:
Given:
-x^5+2x^3+x-1
First separate out the scalar factor
-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
That means that the only possible rational zeros are
We find:
f(1) = 1-2-1+1 = -1
f(-1) = -1+2+1+1 = 3
So
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)
" "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
Notice the remainder is:
(a^4+a^2-2)x+(1-a^3)
We can make this
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)