2007 iTest Problems/Problem TB2

Problem

Factor completely over integer coefficients the polynomial $p(x)=x^8+x^5+x^4+x^3+x+1$ . Demonstrate that your factorization is complete.

Solution

Note that $p(x)=(x^5+x+1)+(x^8+x^4+x^3)$ . If $x\neq 1$ and $x^3=1$ , then $x^5+x+1=x^2+x+1=0$ and $x^8+x^4+x^3=x^2+x+1=0$ . Therefore if $x\neq 1$ and $x^3=1$ , then $p(x)=0$ . Hence $x^2+x+1|p(x)$ . Dividing through gives us

$p(x)=(x^2+x+1)(x^6-x^5+2x^3-x^2+1)$

Using the Rational Root Theorem on the second polynomial gives us that $\pm 1$ are possible roots. Only $-1$ is a possible root. Dividing through gives us

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

Note that $x^5-2x^4+2x^3-x+1$ can be factored into the product of a cubic and a quadratic. Let the product be

$x^5-2x^4+2x^3-x+1=(x^2+ax+b)(x^3+cx^2+dx+e)$

We would want the coefficients to be integers, hence we shall only look for integer solutions. The following equations must then be satisfied:

$be=1$
$ae+bd=-1$
$c+ad+e=0$
$b+ac+d=2$
$a+c=-2$

Since $b$ and $e$ are integers, $(b,e)$ is either $(1,1)$ or $(-1,-1)$ . Testing the first one gives

$a+d=-1$
$c+ad=-1$
$ac+d=1$
$a+c=-2$

We must have that $(a+c)-(a+d)=c-d=-2--1=-1$ $(ac+d)-(c+ad)=a(c-d)-(c-d)=(a-1)(c-d)=1--1=2$ . Therefore $a-1=-2$ , or $a=-1$ . Solving for $c$ and $d$ gives $(a,b,c,d,e)=(-1,1,-1,0,1)$ . We don't need to test the other one.

Hence we have

$p(x)=(x+1)(x^2+x+1)(x^2-x+1)(x^3-x^2+1)$

For any of the factors of degree more than 1 to be factorable in the integers, they must have rational roots, since their degrees are less than 4. None of them have rational roots. Hence $p(x)$ is completely factored.

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2007 iTest Problems/Problem TB2

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