| Version 8 |
Version 7 |
| Factoring a given polynomial may in certain special cases \PMlinkescapetext{succeed} by using the following {\em grouping method}: |
Factoring a given polynomial may in certain special cases \PMlinkescapetext{succeed} by using the following {\em grouping method}: |
| \begin{enumerate} |
\begin{enumerate} |
| \item \PMlinkescapetext{Group the terms of the polynomial in two (or sometimes more) suitable groups}. |
\item \PMlinkescapetext{Group the terms of the polynomial in two (or sometimes more) suitable groups}. |
| \item Factorize the \PMlinkescapetext{groups} separately. |
\item Factorize the \PMlinkescapetext{groups} separately. |
| \item The whole polynomial may then possibly be written in form of a product. |
\item The whole polynomial may then possibly be written in form of a product. |
| \end{enumerate} |
\end{enumerate} |
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| \textbf{Examples} |
\textbf{Examples} |
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| a) \,\,$x^3-x^2-x+1 = \{x^3-x^2\}+\{-x+1\} = x^2(x-1)-1(x-1) = (x-1)(x^2-1)\\ = (x-1)^2(x+1)$ |
a) \,\,$x^3-x^2-x+1 = \{x^3-x^2\}+\{-x+1\} = x^2(x-1)-1(x-1) = (x-1)(x^2-1)\\ = (x-1)^2(x+1)$ |
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| b) \,\,$x^4+3x^3-3x-1 = \{x^4-1\}+\{3x^3-3x\} = (x^2+1)(x^2-1)+3x(x^2-1)\\ = |
b) \,\,$x^4+3x^3-3x-1 = \{x^4-1\}+\{3x^3-3x\} = (x^2+1)(x^2-1)+3x(x^2-1)\\ = |
| (x^2-1)(x^2+1+3x) = (x-1)(x+1)(x^2+3x+1)$ |
(x^2-1)(x^2+1+3x) = (x-1)(x+1)(x^2+3x+1)$ |
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| c) \,\,$x^4+4 = \{x^4+4x^2+4\}-4x^2 = (x^2+2)^2-(2x)^2 = (x^2+2+2x)(x^2+2-2x)\\ = (x^2+2x+2)(x^2-2x+2)$ |
c) \,\,$x^4+4 = \{x^4+4x^2+4\}-4x^2 = (x^2+2)^2-(2x)^2 = (x^2+2+2x)(x^2+2-2x)\\ = (x^2+2x+2)(x^2-2x+2)$ |
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| d) \,\,$x^4+x^2+1 = \{x^4+2x^2+1\}-x^2 = (x^2+1)^2-x^2 = (x^2+1+x)(x^2+1-x)\\ |
d) \,\,$x^4+x^2+1 = \{x^4+2x^2+1\}-x^2 = (x^2+1)^2-x^2 = (x^2+1+x)(x^2+1-x)\\ |
| = (x^2+x+1)(x^2-x+1)$ |
= (x^2+x+1)(x^2-x+1)$ |
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| The trinomials $x^2\!+\!3x\!+\!1$, $x^2\!\pm\!2x\!+\!2$ and $x^2\!\pm\!x\!+\!1$ are irreducible polynomials. |
The trinomials $x^2\!+\!3x\!+\!1$, $x^2\!\pm\!2x\!+\!2$ and $x^2\!\pm\!x\!+\!1$ are irreducible polynomials. |
