How do you factor P(x) = x^5 - x^4 + 2 x^3 + 1?

If it had rational roots, they'd have to either be 1 or -1, but a quick check verifies that neither one works.

So if this factors nicely, it will be a product of a quadratic and a cubic. And if it factors nicely, the lead coefficients will be ones, as are the constant coefficients (ones or negative ones, actually).

So I'm left looking at (x^2 + ax + 1)(x^3 + bx^2 + cx + 1) or (x^2 + ax - 1)(x^3 + bx^2 + cx - 1) as possible factorization patterns.

(x^2 + ax + 1)(x^3 + bx^2 + cx + 1) gives x^5 + (a+b)x^4 + (1+ab+c)x^3 + (b+ac+1)x^2 + (a+c)x+1

So a+c=0, c = -a. And b+ac+1=0, so b-a^2+1=0, b = a^2 - 1. a+b=-1, so a^2 + a - 1 = -1, and a^2+a=0, so either a=0 or a = -1.

If a=0, then we get a contradiction (c=-a=0, and yet our x^3 term would want c=1 in this case.

So a = -1, c = 1, and for the x^4 term to work, b=0. A quick check shows the other terms work out here, and we have a factorization!

x^5-x^4+2x^3+1 = (x^2 - x + 1)(x^3 + x + 1)

x^2-x+1 is irreducible (its roots are complex). x^3+x+1 should have one real root, but by the rational root theorem we know it is an irrational root. So if we're looking for factorizations over the integers, we're done.

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Looking back, I see there was an alternate path we didn't try, guessing at a factorization with constant terms -1. Let's see where that path goes:

(x^2 + ax - 1)(x^3 + bx^2 + cx - 1) = x^5 + (a+b)x^4 + (c+ab-1)x^3 + (-b+ac-1)x^2 + (-a-c)x + 1

So -a-c=0, so c = -a. Now -b+ac-1=0, so b = -a^2-1. And a+b=-1, so -a^2+a-1=-1, so -a^2+a=0, and either a=0 or a=1. If a = 0, then c = 0, and b = -1. But then c+ ab - 1 = -1, not 2, so that doesn't work. And if a = 1, then c = -1 and b = -2, but then c+ab-1 = -4 not 2, so that doesn't work either.

Thus this possible path to a factorization is a dead end.

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Is there a simpler approach? Is this what you had in mind?

First, I would graph the equation. Since it crosses the x-axis only once, it must have four complex roots. That means it can be factored as (x + a) ( x^2 + bx + c) ( x^2 + dx + e). The constants a, b, c, d, and e are all real numbers.

Now, the exact form form the constant "a" involves the cube roots of square root forms; I don't see any easy way to get that. The numeric value to 6 or so decimal places is relatively easy to get, if you know how to differentiate and Newton's method. Frankly, that seems to me to be moderately advanced high school math, beyond Algebra II.

(x-1) and (x+1) are not factors.

It takes some guess work, but it can be factored as (x^2-x+1) ( x^3 +x + 1). The first term has complex factors, and the last term factors as (x + a) ( x^2 -ax + 1/a), with a = 0.682328.

Overall, I would rate this as "very difficult" at the 12th grade level, and "extremely difficult at the 11th grade and below.

Instead of starting with the polynomial, start with the answer. Make the polynomial in factored form then expand it out. This way you can keep the roots real. 4 complex roots is a bit harsh for high school students. It's a bit harsh for undergrads.

p(x)=x^5-x^4+2x^3+1 has 1 real root &

4 non-real roots, which are

x=-0.6823278

x=0.34115+/-1.161519i

x=0.5+/-0.8660572i

Note that not all polynomial p(x) can be factored

exactly. In this example, we can only find the app-

roximate roots only for practical use. Theorectically

if the roots are found we can write down the factors

for p(x) (approximately).

These results are worked out by my own software.

There is no general solution to a 5th or higher order equation, but I'd use the Eisenstanz criterion and start with either (x-1) or (x+1) and see what I got.