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2007 iTest Problems/Problem TB2 - Revision history
2024-03-29T14:18:19Z
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Piphi: /* See also */
2020-01-03T00:05:05Z
<p><span dir="auto"><span class="autocomment">See also</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:05, 3 January 2020</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==See also==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==See also==</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">{{iTest box|year=2007|num-b=TB1|num-a=TB3}}</ins></div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Intermediate Algebra Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Intermediate Algebra Problems]]</div></td></tr>
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Piphi
https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_TB2&diff=53003&oldid=prev
Naysh: /* Solution */
2013-06-13T21:28:07Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 21:28, 13 June 2013</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 <math>p(x)</math> is completely factored.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>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 <math>p(x)</math> is completely factored.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">===Alternate Solution===</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">We write <cmath>p(x)=x^8+x^5+x^4+x^3+x+1=</cmath> <cmath>x^5(x^3+1)+x^3(x+1)+(x+1)=</cmath> <cmath>(x+1)(x^5(x^2-x+1) + (x^3+1))=</cmath> <cmath>(x+1)(x^5(x^2-x+1)+(x+1)(x^2-x+1))=</cmath> <cmath>(x+1)(x^2-x+1)(x^5+x+1)=</cmath> <cmath>(x+1)(x^2-x+1)(x^2+x+1)(x^3-x^2+1)</cmath> </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The factorization of <math>x^5+x+1</math> is trivial once we look at the exponents modulo <math>3</math>; since any root <math>\omega</math> of <math>x^2+x+1</math> satisfies <math>\omega^3=1</math>, it follows that <math>x^2+x+1 | x^5+x+1</math> and the cubic factor comes as a result of polynomial division.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">To prove that this is a complete factorization, it suffices to note that the factors of degree greater than <math>1</math> have no rational roots (follows from the rational root theorem and some small cases).</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
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Naysh
https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_TB2&diff=31642&oldid=prev
1=2: This problem was fun.
2009-05-14T16:42:17Z
<p>This problem was fun.</p>
<p><b>New page</b></p><div>== Problem ==<br />
Factor completely over integer coefficients the polynomial <math>p(x)=x^8+x^5+x^4+x^3+x+1</math>. Demonstrate that your factorization is complete.<br />
<br />
== Solution ==<br />
Note that <math>p(x)=(x^5+x+1)+(x^8+x^4+x^3)</math>. If <math>x\neq 1</math> and <math>x^3=1</math>, then <math>x^5+x+1=x^2+x+1=0</math> and <math>x^8+x^4+x^3=x^2+x+1=0</math>. Therefore if <math>x\neq 1</math> and <math>x^3=1</math>, then <math>p(x)=0</math>. Hence <math>x^2+x+1|p(x)</math>. Dividing through gives us<br />
<br />
<center><math>p(x)=(x^2+x+1)(x^6-x^5+2x^3-x^2+1)</math></center><br />
<br />
Using the [[Rational Root Theorem]] on the second polynomial gives us that <math>\pm 1</math> are possible roots. Only <math>-1</math> is a possible root. Dividing through gives us<br />
<br />
<center><math>p(x)=(x+1)(x^2+x+1)(x^5-2x^4+2x^3-x+1)</math></center><br />
<br />
Note that <math>x^5-2x^4+2x^3-x+1</math> can be factored into the product of a cubic and a quadratic. Let the product be<br />
<br />
<center><math>x^5-2x^4+2x^3-x+1=(x^2+ax+b)(x^3+cx^2+dx+e)</math></center><br />
<br />
We would want the coefficients to be integers, hence we shall only look for integer solutions. The following equations must then be satisfied:<br />
* <math>be=1</math><br />
* <math>ae+bd=-1</math><br />
* <math>c+ad+e=0</math><br />
* <math>b+ac+d=2</math><br />
* <math>a+c=-2</math><br />
Since <math>b</math> and <math>e</math> are integers, <math>(b,e)</math> is either <math>(1,1)</math> or <math>(-1,-1)</math>. Testing the first one gives<br />
* <math>a+d=-1</math><br />
* <math>c+ad=-1</math><br />
* <math>ac+d=1</math><br />
* <math>a+c=-2</math><br />
We must have that <math>(a+c)-(a+d)=c-d=-2--1=-1</math> <math>(ac+d)-(c+ad)=a(c-d)-(c-d)=(a-1)(c-d)=1--1=2</math>. Therefore <math>a-1=-2</math>, or <math>a=-1</math>. Solving for <math>c</math> and <math>d</math> gives <math>(a,b,c,d,e)=(-1,1,-1,0,1)</math>. We don't need to test the other one.<br />
<br />
Hence we have<br />
<br />
<center><math>p(x)=(x+1)(x^2+x+1)(x^2-x+1)(x^3-x^2+1)</math></center><br />
<br />
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 <math>p(x)</math> is completely factored.<br />
<br />
==See also==<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>
1=2