【数学I】例題1.1.12：因数分解の工夫（次数が同じ場合）（One More）★★

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% 例題I1.1.12：因数分解の工夫（次数が同じ場合）（One More）★★
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次の式を因数分解せよ．

(1) $x^2+3xy+2y^2-x-3y-2$

(2) $2x^2+5xy+3y^2-3x-5y-2$

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% 例題I1.1.12の解答（One More）★★
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(1)

\begin{align*}
&x^2+3xy+2y^2-x-3y-2\\
=&x^2+(3y-1)x+(2y^2-3y-2)\\
=&x^2+(3y-1)x+(y-2)(2y+1)\cdots(\mathrm{i})\\
=&\{x+(y-2)\}\{x+(2y+1)\}\cdots(\mathrm{ii})\\
=&(x+y-2)(x+2y+1)
\end{align*}

(i)

(ii)

(2)

\begin{align*}
&2x^2+5xy+3y^2-3x-5y-2\\
=&2x^2+(5y-3)x+(3y^2-5y-2)\\
=&2x^2+(5y-3)x+(y-2)(3y+1)\cdots(\mathrm{i})\\
=&\{x+(y-2)\}\{2x+(3y+1)\}\cdots(\mathrm{ii})\\
=&(x+y-2)(2x+3y+1)
\end{align*}

(i)

(ii)

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% 問題I1.1.12：因数分解の工夫（次数が同じ場合）（One More）★★
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次の式を因数分解せよ．

(1) $x^2+4xy+3y^2-2x-8y-3$

(2) $3x^2+11xy+10y^2-x-3y-4$

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% 問題I1.1.12の解答（One More）★★
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(1)

\begin{align*}
&x^2+4xy+3y^2-2x-8y-3\\
=&x^2+(4y-2)x+(3y^2-8y-3)\\
=&x^2+(4y-2)x+(y-3)(3y+1)\cdots(\mathrm{i})\\
=&\{x+(y-3)\}\{x+(3y+1)\}\cdots(\mathrm{ii})\\
=&(x+y-3)(x+3y+1)
\end{align*}

(i)

(ii)

(2)

\begin{align*}
&3x^2+11xy+10y^2-x-3y-4\\
=&3x^2+(11y-1)x+(10y^2-3y-4)\\
=&3x^2+(11y-1)x+(2y+1)(5y-4)\cdots(\mathrm{i})\\
=&\{x+(2y+1)\}\{3x+(5y-4)\}\cdots(\mathrm{ii})\\
=&(x+2y+1)(3x+5y-4)
\end{align*}

(i)

(ii)

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% 例題I1.1.12の側注：因数分解の工夫（次数が同じ場合）（One More）★★
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% 側注1
先に，(i)のように$2y^2-3y-2$をたすき掛けを用いて因数分解する．次に，(ii)のように全体を$x$の2次式と考えて，たすき掛けを用いて因数分解する．

% 側注2
$y$について整理して，因数分解してもよい．

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% 問題I1.1.12の側注：因数分解の工夫（次数が同じ場合）（One More）★★
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% 側注1
先に，(i)のように$3y^2-8y-3$をたすき掛けを用いて因数分解する．次に，(ii)のように全体を$x$の2次式と考えて，たすき掛けを用いて因数分解する．

% 側注2
$y$について整理して，因数分解してもよい．

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% 例題I1.1.12の考え方・補足など：因数分解の工夫（次数が同じ場合）（One More）★★
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% 考え方
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複数の文字を含む式の因数分解は，次数の低い文字に着目するとよい．しかし，(1)，(2)のように，$x,y$どちらについても次数が同じ場合がある（$x,y$ともに2次式である）．このような場合は，どちらかの文字で整理すればよい．なお，解答において(1)は，$x^2$の係数が1であることから$x$について整理している（計算が楽になることが見込まれる）．