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!