Process of thinking whe factoring x^35-x^19-x^17-x^2+1

Question

Can someone explain their process of thinking when factoring this expression : x^35-x^19-x^17-x^2+1  ?

Answer 1

Can someone explain their process of thinking when factoring this expression : x^35-x^19-x^17-x^2+1  ?

There are various methods for factoring an expression, but these usually involve linear factors, and guesswork, based upon past experience.

Also, sometimes an expression of higher order can have its order reduced by finding roots and extracting linear factors one at a time.

There is a procedural method for finding factors (as opposed to roots) for a general polynomial.

Let us start with a quadratic:  x^2 +αx + β = 0, and assume two linear factors (x – a) and (x – b).

Then,

x^2 +αx + β = (x – a)(x – b)

x^2 +αx + β = x^2 + (a+b)x + ab

Equating coefficients of powers of x we get the following simultaneous equations.

a + b = α

ab = β

Solving for b gives us a quadratic in b, viz. b^2 – αb + β = 0.

Using the quadratic formula gives us, b = {α +/- √(α^2 – 4β)) / 2

If we limit ourselves to real solutions (α^2 >= 4β), and rational solutions (α^2 – 4β = γ^2, where γ is a rational number, then we will have two factors.

But what if we now have a larger polynomial that we wish to factor,

e.g. x^20 + a1.x^19 + a2.X^18 + ...

We can assume 20 linear factors and use the previous procedure to find real, rational factors.

If that fails, assume one quadratic factor and 18 other linear factors. Create your set of simultaneous equations and once again look for real rational solutions.

If that fails, repeat with a cubic factor and so on up until x^10. If still no solutions, then the polynomial is unfactorisable.

If you are successful in finding a higher-order factor, then extract that factor from the original polynomial, leaving a polynomial of reduced power, and repeat the above procedure.

Obviously, with large polynomials, it would not be a good idea to do this by hand (you could get a pretty large set of simultaneous equations), but instead use a computer algebra program, such as Mathemtica or Maple. Such applications would use software procedures similar to the above to output a solution for the user.

The actual solution to your polynomial is:

x^35 – x^19 – x^17 – x^2 + 1 = (x^18 + x – 1)(x^17 – x –  1)

(I used Maple)

Answer 2

Let f(x)=x^35-x^19-x^17-x^2+1.

First try x=1 and -1: when x=-1, f(-1)=1 and f(1)=-1, so x±1 is not a factor.

Now assume that there are two factors g(x) and h(x) such that f(x)=g(x) * h(x).

Since the coefficient of x^35 is 1, the highest order of x in g(x) is x^a and that in h(x) is x^(35-a).

If we consider the terms -x^2+1 as one unit, we know the factors are (-x+1)(x+1), so f(x) could be written:

f(x)=x^35-x^19-x^17+(-x+1)(x+1), reducing the terms from 5 to 4.

Therefore, (a) g(x)=x^a+p(x)+(-x+1) and h(x)=x^(35-a)+q(x)+(x+1) or:

(b) g(x)=x^a+p(x)-(-x+1) and h(x)=x^(35-a)+q(x)-(x+1), where p and q are functions to be determined.

Consider (a) first.

First consider the case where there is no intermediate term, p(x)=q(x)=0, so g(x)=x^a-x+1 and h(x)=x^(35-a)+x+1.

g(x)*h(x)=x^35-x^2+1+(x+1)x^a+(-x+1)x^(35-a).

So (x+1)x^a+(-x+1)x^(35-a)=x^(a+1)+x^a-x^(36-a)+x^(35-a) must be equivalent to -x^19-x^17. Other terms must cancel out. But there is only one negative term and we need two. There is no solution on this path.

Now consider (b), and put p(x)=q(x)=0 again:

g(x)*h(x)=x^35-x^2+1-(x+1)x^a-(-x+1)x^(35-a)=x^35-x^2+1-(x+1)x^a+(x-1)x^(35-a).

This time -(x+1)x^a+(x-1)x^(35-a)=-x^(a+1)-x^a+x^(36-a)-x^(35-a) must be equivalent to -x^19-x^17. Other terms must cancel out. x^(a+1)+x^a-x^(36-a)+x^(35-a)=x^19+x^17. There is only one negative term, -x^(36-a), so it must be the reverse of one of three positive terms. The only term it can be is x^a, so equating exponents, a=36-a, 2a=36, a=18.

The solution is therefore (b) (x^18-(-x+1))(x^17-(x+1))=(x^18+x-1)(x^17-x-1).

Having discovered a solution when p and q are both zero, it is unnecessary to seek binary factor solutions for non-zero p and q. Also, we cannot factorise x^18+x-1 to obtain irrational factors of the form g(x)=x^a±(√x-1), h(x)=x^(18-a)±(√x+1), because g*h(x) would produce x^18+x-1±x^a(√x+1)±x^(18-a)(√x-1)=[x^18+x-1]±x^(a+1/2)±x^a±x^(18.5-a)Tx^(18-a). This residual expression cannot be zero, which would be a requirement, since we already have x^18+x-1.