【数学I】例題1.1.16：因数分解（ax^4+bx^2+cの形）（One More）★★★

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% 例題I1.1.16：因数分解（$ax^4+bx^2+c$の形）（One More）★★★
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次の式を因数分解せよ．

(1) $x^4-7x^2+12$

(2) $x^4+2x^2+9$

(3) $x^4-15x^2y^2+9y^4$

(4) $4x^4+1$

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% 例題I1.1.16の解答（One More）★★★
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(1) $x^4-7x^2+12=(x^2-4)(x^2-3)=(x+2)(x-2)(x^2-3)$

(2)

\begin{align*}
x^4+2x^2+9&=(x^4+6x^2+9)-4x^2\\
&=(x^2+3)^2-(2x)^2\\
&=\{(x^2+3)+2x\}\{(x^2+3)-2x\}\\
&=(x^2+2x+3)(x^2-2x+3)
\end{align*}

(3)

\begin{align*}
x^4-15x^2y^2+9y^4&=(x^4-6x^2y^2+9y^4)-9x^2y^2\\
&=(x^2-3y^2)^2-9x^2y^2\\
&=\{(x^2-3y^2)+3xy\}\{(x^2-3y^2)-3xy\}\\
&=(x^2+3xy-3y^2)(x^2-3xy-3y^2)
\end{align*}

(4)

\begin{align*}
4x^4+1&=(4x^4+4x^2+1)-4x^2\\
&=(2x^2+1)^2-(2x)^2\\
&=\{(2x^2+1)+2x\}\{(2x^2+1)-2x\}\\
&=(2x^2+2x+1)(2x^2-2x+1)
\end{align*}

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% 問題I1.1.16：因数分解（$ax^4+bx^2+c$の形）（One More）★★★
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次の式を因数分解せよ．

(1) $x^4-11x^2+18$

(2) $x^4+4x^2+16$

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

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% 問題I1.1.16の解答（One More）★★★
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(1) $x^4-11x^2+18=(x^2-9)(x^2-2)=(x+3)(x-3)(x^2-2)$

(2)

\begin{align*}
x^4+4x^2+16&=(x^4+8x^2+16)-4x^2\\
&={(x^2+4)^2}-(2x)^2\\
&=\{(x^2+4)+2x\}\{(x^2+4)-2x\}\\
&=(x^2+2x+4)(x^2-2x+4)
\end{align*}

(3)

\begin{align*}
x^4-8x^2y^2+4y^4&=(x^4-4x^2y^2+4y^4)-4x^2y^2\\
&={(x^2-2y^2)^2}-4x^2y^2\\
&=\{(x^2-2y^2)+2xy\}\{(x^2-2y^2)-2xy\}\\
&=(x^2+2xy-2y^2)(x^2-2xy-2y^2)
\end{align*}

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% 例題I1.1.16の側注：因数分解（$ax^4+bx^2+c$の形）（One More）★★★
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% 側注1
$x^2=A$とおくと，

\[
A^2-7A+12=(A-4)(A-3)
\]

% 側注2
$x^4$と定数項$9$より，$(x^2+3)^2$または$(x^2-3)^2$を作ることを考える．

% 側注3
$x^4$と$9y^4$より，$(x^2+3y^2)^2$または$(x^2-3y^2)^2$を作ることを考える．

% 側注4
$4x^4$と定数項$1$より，$(2x^2+1)^2$または$(2x^2-1)^2$を作ることを考える．

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% 問題I1.1.16の側注：因数分解（$ax^4+bx^2+c$の形）（One More）★★★
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% 側注1
$x^2=A$とおくと，

\[
A^2-11A+18=(A-9)(A-2)
\]

% 側注2
$x^4$と定数項$16$より，$(x^2+4)^2$または$(x^2-4)^2$を作ることを考える．

% 側注3
$x^4$と$4y^4$より，$(x^2+2y^2)^2$または$(x^2-2y^2)^2$を作ることを考える．

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% 例題I1.1.16の考え方・補足など：因数分解（$ax^4+bx^2+c$の形）（One More）★★★
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% 考え方
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$ax^4+bx^2+c$の形の式を，複2次式という（各項の次数が偶数の4次式のことを指すことが多いが，次数が偶数の6次以上の式を指すこともある）．$x^2=A$とおき，$aA^2+bA+c$の形に変形したとき，公式を用いた因数分解が直接できない場合は，平方の差の形を作ることを考える．

(1) $x^2=A$とおくと，$A^2-7A+12$となり，因数分解の公式$x^2+(a+b)x+ab=(x+a)(x+b)$を使うことができる．

(2) $x^2=A$とおいても，$A^2+2A+9$となり，直接公式を用いた因数分解ができない．そこで，$x^4$と定数項$9$に注目して，$(x^2+3)^2$または$(x^2-3)^2$を作ることを考える．ここでは$(x^2+3)^2$を作れば，平方の差の形に変形することができる．

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\begin{align*}
ax^4+bx^2+c\text{の形の因数分解}&\rightarrow x^2=A\text{とおき換える．}\\
&\rightarrow\text{公式を用いた因数分解が直接できないときは，平方の差を作る．}
\end{align*}

\[
x^4+2x^2+\underbrace{9}_{3\times 3}=x^4+\underbrace{6}_{2\times 3}x^2+\underbrace{9}_{3\times 3}-4x^2=(x^2+3)^2-(2x)^2
\]