How many prime factors are obtained? x^7 + c^3x^4 - c^4x^3-c^7

x^7 + c^3x^4 - c^4x^3-c^7

1 Answer
May 9, 2018

x^7+c^3x^4-c^4x^3-c^7=(x^2-cx+c^2)(x^2+c^2)(x+c)^2(x-c)

Explanation:

We are asked to factor

x^7+c^3x^4-c^4x^3-c^7

Not so easy this one.

First note that if x=c then the expression yields zero.

c^7+c^3c^4-c^4c^3-c^7=c^7+c^7-c^7-c^7=0

This means that (x-c) is a factor of x^7+c^3x^4-c^4x^3-c^7.

Let's do the long division and see what remains.

x^7+c^3x^4-c^4x^3-c^7

=x^7-cx^6+cx^6-c^2x^5+c^2x^5-c^3x^4+2c^3x^4-2c^4x^3+c^4x^3-c^5x^2+c^5x^2-c^6x+c^6x-c^7

=x^6(x-c)+cx^5(x-c)+c^2x^4(x-c)+2c^3x^3(x-c)+c^4x^2(x-c)+c^5x(x-c)+c^6(x-c)

=(x^6+cx^5+c^2x^4+2c^3x^3+c^4x^2+c^5x+c^6)(x-c)

Now we notice that when x=-c,

x^6+cx^5+c^2x^4+2c^3x^3+c^4x^2+c^5x+c^6=c^6-c^6+c^6-2c^6+c^6-c^6+c^6=0

Let's divide x^6+cx^5+c^2x^4+2c^3x^3+c^4x^2+c^5x+c^6 by (x+c).

x^6+cx^5+c^2x^4+2c^3x^3+c^4x^2+c^5x+c^6

=x^6+cx^5+c^2x^4+c^3x^3+c^3x^3+c^4x^2+c^5x+c^6

=x^5(x+c)+c^2x^3(x+c)+c^3x^2(x+c)+c^5(x+c)

=(x^5+c^2x^3+c^3x^2+c^5)(x+c)

Next notice that when x=-c

x^5+c^2x^3+c^3x^2+c^5=-c^5-c^5+c^5+c^5=0

So lets divide x^5+c^2x^3+c^3x^2+c^5 by (x+c).

x^5+c^2x^3+c^3x^2+c^5

=x^5+cx^4-cx^4-c^2x^3+2c^2x^3+2c^3x^2-c^3x^2-c^4x+c^4x+c^5

=x^4(x+c)-cx^3(x+c)+2c^2x^2(x+c)-c^3x(x+c)+c^4(x+c)

=(x^4-cx^3+2c^2x^2-c^3x+c^4)(x+c)

Now we are forced to factor x^4-cx^3+2c^2x^2-c^3x+c^4.

A little "inspired grouping" gets us over this hurdle.

Start with

x^4-cx^3+2c^2x^2-c^3x+c^4.

Split the third term.

x^4-cx^3+c^2x^2+c^2x^2-c^3x+c^4

Commute the 3rd term.

x^4+c^2x^2-cx^3-c^3x+c^2x^2+c^4

Group and factor.

x^2(x^2+c^2)-cx(x^2+c^2)+c^2(x^2+c^2)

(x^2-cx+c^2)(x^2+c^2)

If c and x are both real numbers, then we cannot factor any further. Note that the only way for x^2+c^2=0 is if they are both zero or if either x^2 or c^2 are less than zero. The only way for to have a squared number be less than zero is if it is imaginary. Also note that by using the quadratic formula, we can find x's that satisfy x^2-cx+c^2=0.

x=(cpmsqrt(c^2-4c^2))/2=(cpmcsqrt(-3))/2

So both roots for this expression MUST be imaginary regardless of whether c is imaginary or not.

So we have the factored form with the following logic:

x^7+c^3x^4-c^4x^3-c^7

=(x^6+cx^5+c^2x^4+2c^3x^3+c^4x^2+c^5x+c^6)(x-c)

=(x^5+c^2x^3+c^3x^2+c^5)(x+c)(x-c)

=(x^4-cx^3+2c^2x^2-c^3x+c^4)(x+c)(x+c)(x-c)

=(x^2-cx+c^2)(x^2+c^2)(x+c)(x+c)(x-c)

=(x^2-cx+c^2)(x^2+c^2)(x+c)^2(x-c)

Yikes! Your teacher sure is mean!