MathJax快速範本-輾轉相除法

$設 f(x)=x^5-x^3+2x^2-2x-4，g(x)=x^4+x^3+x^2+3x+2，\\ h(x) 為f(x)與g(x)的最高公因式且最高次項係數為1，則h(1) 與 h(2)的乘積為?$ 方法很多，本題比較適合用輾轉相除法(當兩式次方相近時最好用)

$$\begin{array}{c|l|l|c} x & x^5+0x^4-x^3+2x^2-2x-4 & x^4+~x^3+x^2+3x+2 & -1\\    &  x^5+x^4+x^3+3x^2+2x &  x^4-0x^3+x^2+2x &  \\ \hline -1 & 0x^5-x^4-2x^3-x^2-4x-4 & 0x^4+x^3+0x^2+x+2 & -1 \\  & 0x^5-x^4-x^3-x^2-3x-2 & 0x^4+x^3+0x^2+x+2 & \\ \hline & 0x^5+0x^4-x^3+0x^2-x-2 & & \end{array}$$  

$故最高公因式為h(x)=x^{3}+x+2$

 $故h(1)× h(2)=(1+1+2)(8+2+2)=48$

 修改後的版本

$$\left. \begin{gathered} \hfill \underline {\begin{array}{*{20}{c}} {} \\ { + x} \end{array}} \\ \hfill \underline {\begin{array}{*{20}{c}} {} \\ { - 1} \end{array}} \\ \hfill \begin{array}{*{20}{c}} {} \\ {} \end{array} \\ \end{gathered} \right|\left. \begin{gathered} \hfill \underline {\begin{array}{*{20}{c}} {{x^5}}&{ + 0{x^4}}&{ - {x^3}}&{ + 2{x^2}}&{ - 2x}&{ - 4} \\ {{x^5}}&{ + {x^4}}&{ + {x^3}}&{ + 3{x^2}}&{ + 2x}&{} \end{array}} \\ \hfill \underline {\begin{array}{*{20}{c}} {}&{ - {x^4}}&{ - 2{x^3}}&{ - {x^2}}&{ - 4x}&{ - 4} \\ {}&{ - {x^4}}&{ - {x^3}}&{ - {x^2}}&{ - 3x}&{ - 2} \end{array}} \\ \hfill \begin{array}{*{20}{c}} {}&{}&{ - {x^3}}&{ + 0{x^2}}&{ - x}&{ - 2} \\ {}&{}&{}&{}&{}&{} \end{array} \\ \end{gathered} \right|\left| \begin{gathered} \hfill \underline {\begin{array}{*{20}{c}} {{x^4}}&{ + {x^3}}&{ + {x^2}}&{ + 3x}&{ + 2} \\ {{x^4}}&{ + 0{x^3}}&{ + {x^2}}&{ + 2x}&{} \end{array}} \\ \hfill \underline {\begin{array}{*{20}{c}} {}&{{x^3}}&{ + 0{x^2}}&{ + x}&{ + 2} \\ {}&{{x^3}}&{ + 0{x^2}}&{ + x}&{ + 2} \end{array}} \\ \hfill \begin{array}{*{20}{c}} {}&{}&{}&{}&{} \\ {}&{}&{}&{}&{} \end{array} \\ \end{gathered} \right.\left| \begin{gathered} \hfill \underline {\begin{array}{*{20}{c}} {} \\ { - 1} \end{array}} \\ \hfill \underline {\begin{array}{*{20}{c}} {} \\ { - 1} \end{array}} \\ \hfill \begin{array}{*{20}{c}} {} \\ {} \end{array} \\ \end{gathered} \right.$$