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【数学I】例題1.1.10:因数分解(3次式)(One More)★★

【数学I】例題1.1.10:因数分解(3次式)(One More)
【数学I】例題1.1.10:因数分解(3次式)の例題ページ
問題の解答

【数学I】問題1.1.10:因数分解(3次式)の解答
検索用コード(LaTeX)
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次の式を因数分解せよ.

(1) $x^3+64$

(2) $8a^3-125b^3$

(3) $x^3+3x^2+3x+1$

(4) $x^3+2x^2-9x-18$

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(1)

\begin{align*}
x^3+64&=x^3+4^3\\
&=(x+4)\{x^2-x\cdot 4+4^2\}\\
&=(x+4)(x^2-4x+16)
\end{align*}

(2)

\begin{align*}
8a^3-125b^3&=(2a)^3-(5b)^3\\
&=(2a-5b)\{(2a)^2+2a\cdot 5b+(5b)^2\}\\
&=(2a-5b)(4a^2+10ab+25b^2)
\end{align*}

(3) $x^3+3x^2+3x+1=(x+1)^3$

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\begin{align*}
x^3+3x^2+3x+1&=(x^3+1)+(3x^2+3x)\\
&=(x+1)(x^2-x+1)+3x(x+1)\\
&=(x+1)\{(x^2-x+1)+3x\}\\
&=(x+1)(x^2+2x+1)\\
&=(x+1)(x+1)^2\\
&=(x+1)^3
\end{align*}

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(4)

\begin{align*}
x^3+2x^2-9x-18&=(x^3+2x^2)-(9x+18)\\
&=x^2(x+2)-9(x+2)\\
&=(x+2)(x^2-9)\\
&=(x+2)(x-3)(x+3)
\end{align*}

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% 問題I1.1.10:因数分解(3次式)(One More)★★
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次の式を因数分解せよ.

(1) $x^3-27$

(2) $27m^3+8n^3$

(3) $x^3-6x^2+12x-8$

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

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(1)

\begin{align*}
x^3-27&=x^3-3^3\\
&=(x-3)\{x^2+x\cdot 3+3^2\}\\
&=(x-3)(x^2+3x+9)
\end{align*}

(2)

\begin{align*}
27m^3+8n^3&=(3m)^3+(2n)^3\\
&=(3m+2n)\{(3m)^2-3m\cdot 2n+(2n)^2\}\\
&=(3m+2n)(9m^2-6mn+4n^2)
\end{align*}

(3) $x^3-6x^2+12x-8=(x-2)^3$

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\begin{align*}
x^3-6x^2+12x-8&=(x^3-8)+(-6x^2+12x)\\
&=(x-2)(x^2+2x+4)-6x(x-2)\\
&=(x-2)\{x^2+2x+4-6x\}\\
&=(x-2)(x^2-4x+4)\\
&=(x-2)(x-2)^2\\
&=(x-2)^3
\end{align*}

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% 問題I1.1.10の解答(One More)★★
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(4)

\begin{align*}
x^3-4x^2-x+4&=(x^3-4x^2)-(x-4)\\
&=x^2(x-4)-1(x-4)\\
&=(x-4)(x^2-1)\\
&=(x-4)(x-1)(x+1)
\end{align*}

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