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		<id>https://en.formulasearchengine.com/w/index.php?title=Nernst_effect&amp;diff=9514</id>
		<title>Nernst effect</title>
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		<updated>2013-04-15T13:19:44Z</updated>

		<summary type="html">&lt;p&gt;120.61.50.47: &lt;/p&gt;
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&lt;div&gt;In [[algebra]], the &#039;&#039;&#039;factor theorem&#039;&#039;&#039; is a theorem linking factors and [[Zero (complex analysis)|zeros]] of a [[polynomial]]. It is a [[special case]] of the [[polynomial remainder theorem]].&amp;lt;ref&amp;gt;{{citation|first=Michael|last=Sullivan|title=Algebra and Trigonometry|page=381|publisher=Prentice Hall|year=1996|isbn=0-13-370149-2}}.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The factor theorem states that a polynomial &amp;lt;math&amp;gt;f(x)&amp;lt;/math&amp;gt; has a factor &amp;lt;math&amp;gt;(x - k)&amp;lt;/math&amp;gt; [[if and only if]] &amp;lt;math&amp;gt;f(k)=0&amp;lt;/math&amp;gt; (i.e. &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; is a root).&amp;lt;ref&amp;gt;{{citation|first1=V K|last1=Sehgal|first2=Sonal|last2=Gupta|title=Longman ICSE Mathematics Class 10|page=119|publisher=Dorling Kindersley (India)|isbn=978-81-317-2816-1}}.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Factorization of polynomials==&lt;br /&gt;
{{Main|Factorization of polynomials}}&lt;br /&gt;
Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.&lt;br /&gt;
&lt;br /&gt;
The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:&amp;lt;ref&amp;gt;{{citation|first=R. K.|last=Bansal|title=Comprehensive Mathematics IX|page=142|publisher=Laxmi Publications|isbn=81-7008-629-9}}.&amp;lt;/ref&amp;gt;&lt;br /&gt;
# &amp;quot;Guess&amp;quot; a zero &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; of the polynomial &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt;. (In general, this can be &#039;&#039;very hard&#039;&#039;, but math textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)&lt;br /&gt;
# Use the factor theorem to conclude that &amp;lt;math&amp;gt;(x-a)&amp;lt;/math&amp;gt; is a factor of &amp;lt;math&amp;gt;f(x)&amp;lt;/math&amp;gt;.&lt;br /&gt;
# Compute the polynomial &amp;lt;math&amp;gt; g(x) = f(x) \big/ (x-a) &amp;lt;/math&amp;gt;, for example using [[polynomial long division]] or [[synthetic division]].&lt;br /&gt;
# Conclude that any root &amp;lt;math&amp;gt;x \neq a&amp;lt;/math&amp;gt; of &amp;lt;math&amp;gt;f(x)=0&amp;lt;/math&amp;gt; is a root of &amp;lt;math&amp;gt;g(x)=0&amp;lt;/math&amp;gt;. Since the [[polynomial degree]] of &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; is one less than that of &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt;, it is &amp;quot;simpler&amp;quot; to find the remaining zeros by studying &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
===Example===&lt;br /&gt;
Find the factors at&lt;br /&gt;
: &amp;lt;math&amp;gt;x^3 + 7x^2 + 8x + 2.&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To do this you would use trial and error to find the first x value that causes the expression to equal zero.  To find out if &amp;lt;math&amp;gt;(x - 1)&amp;lt;/math&amp;gt; is a factor, substitute &amp;lt;math&amp;gt;x = 1&amp;lt;/math&amp;gt; into the polynomial above:&lt;br /&gt;
: &amp;lt;math&amp;gt;x^3 + 7x^2 + 8x + 2 = (1)^3 + 7(1)^2 + 8(1) + 2&amp;lt;/math&amp;gt;&lt;br /&gt;
: &amp;lt;math&amp;gt;= 1 + 7 + 8 + 2&amp;lt;/math&amp;gt;&lt;br /&gt;
: &amp;lt;math&amp;gt;= 18.&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
As this is equal to 18 and not 0 this means &amp;lt;math&amp;gt;(x - 1)&amp;lt;/math&amp;gt; is not a factor of &amp;lt;math&amp;gt;x^3 + 7x^2 + 8x + 2&amp;lt;/math&amp;gt;. So, we next try &amp;lt;math&amp;gt;(x + 1)&amp;lt;/math&amp;gt; (substituting &amp;lt;math&amp;gt;x = -1&amp;lt;/math&amp;gt; into the polynomial):&lt;br /&gt;
: &amp;lt;math&amp;gt;(-1)^3 + 7(-1)^2 + 8(-1) + 2.&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This is equal to &amp;lt;math&amp;gt;0&amp;lt;/math&amp;gt;. Therefore &amp;lt;math&amp;gt;x-(-1)&amp;lt;/math&amp;gt;, which is to say &amp;lt;math&amp;gt;x+1&amp;lt;/math&amp;gt;, is a factor, and &amp;lt;math&amp;gt;-1&amp;lt;/math&amp;gt; is a [[Root of a function|root]] of &amp;lt;math&amp;gt;x^3 + 7x^2 + 8x + 2.&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next two roots can be found by algebraically dividing &amp;lt;math&amp;gt;x^3 + 7x^2 + 8x + 2&amp;lt;/math&amp;gt; by &amp;lt;math&amp;gt;(x+1)&amp;lt;/math&amp;gt; to get a quadratic, which can be solved directly, by the factor theorem or by the [[quadratic equation]].&lt;br /&gt;
&lt;br /&gt;
: &amp;lt;math&amp;gt;{x^3 + 7x^2 + 8x + 2 \over x + 1} = x^2 + 6x + 2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
and therefore &amp;lt;math&amp;gt;(x+1)&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;x^2 + 6x + 2&amp;lt;/math&amp;gt; are the factors of &amp;lt;math&amp;gt;x^3 + 7x^2 + 8x + 2.&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Formal version==&lt;br /&gt;
Let &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt; be a one-variable polynomial with coefficients in a commutative ring &amp;lt;math&amp;gt;R&amp;lt;/math&amp;gt;, and let &amp;lt;math&amp;gt;a \in R&amp;lt;/math&amp;gt;. Then &amp;lt;math&amp;gt;f(a) = 0&amp;lt;/math&amp;gt; if and only if &amp;lt;math&amp;gt;f(x)=(x-a)g(x)&amp;lt;/math&amp;gt; for some polynomial &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt;.   In this case, &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; is determined uniquely.&lt;br /&gt;
&lt;br /&gt;
As for the problem of algorithmically finding all roots, if &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt; is given and &amp;lt;math&amp;gt;a&amp;lt;/math&amp;gt; is known, then &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt; can be computed by [[polynomial long division]]; then one can compute the remaining roots of &amp;lt;math&amp;gt;f&amp;lt;/math&amp;gt;, including repeated roots, by factoring &amp;lt;math&amp;gt;g&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
{{reflist}}&lt;br /&gt;
&lt;br /&gt;
{{DEFAULTSORT:Factor Theorem}}&lt;br /&gt;
[[Category:Polynomials]]&lt;br /&gt;
[[Category:Theorems in algebra]]&lt;/div&gt;</summary>
		<author><name>120.61.50.47</name></author>
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