【数学I】例題1.1.9：たすき掛けを用いた因数分解（One More）★

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次の式を因数分解せよ．

(1) $3x^2+7x+2$

(2) $3x^2-4x-15$

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

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(1) $3x^2+7x+2=(x+2)(3x+1)$

(2) $3x^2-4x-15=(x-3)(3x+5)$

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

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次の式を因数分解せよ．

(1) $2x^2+5x+3$

(2) $4x^2-11x-3$

(3) $6x^2+13xy+6y^2$

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(1) $2x^2+5x+3=(x+1)(2x+3)$

(2) $4x^2-11x-3=(x-3)(4x+1)$

(3) $6x^2+13xy+6y^2=(2x+3y)(3x+2y)$

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% 側注1
（失敗例）

% 側注2
（失敗例）

% 側注3
$y$を忘れないように注意すること．

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% 側注1
$y$を忘れないように注意すること．

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たすき掛けを用いて，$acx^2+(ad+bc)x+bd=(ax+b)(cx+d)$を満たす$a,b,c,d$の組み合わせを考える．なお，たすき掛けの因数分解は，実用的には計算の手間に大差がなかった乗法公式$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$の逆の操作である．

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\[
acx^2+(ad+bc)x+bd=(ax+b)(cx+d)
\]

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例えば(1)は，$3x^2+7x+2$が$(ax+b)(cx+d)$の形に因数分解できるとすると，

\[
3x^2+7x+2=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd
\]

したがって，$ac=3,ad+bc=7,bd=2$を満たす$a,b,c,d$を見つければよい．

\[
ac=3\text{より，}1\times 3,bd=2\text{より，}1\times 2,(-1)\times(-2)
\]

などの分解を考える．下の図式のように，$a,b,c,d$の組み合わせにおいて$ad+bc=7$を満たすものは，$a=1,b=2,c=3,d=1$が該当する（成功するまで，様々な$a,b,c,d$の組み合わせを試す）．

よって，$3x^2+7x+2=(x+2)(3x+1)$と因数分解できる．このような計算を，たすき掛けという．