この記事は，因数分解の演習用ページです．
2次の因数分解の公式で解ける問題の追加分となります．以下の演習問題を解いていない人は，先にこちらを解いてみてください．
また解説ページはこちらです．
演習問題1．
次の式を因数分解せよ．
(1)
x^2 + 3xy + 4x + 2y^2 +7y + 3
(2)
x^2 - 7xy + 5x + 12y^2 -17y + 6
(3)
x^2 + 2xy + 3x - 3y^2 + y + 2
(4)
x^2 - 3xy - 3x + 2y^2 + 5y + 2
(5)
2x^2 - 5xy + 8x + 2y^2 - 7y + 6
(6)
2x^2 - 2xy + x - 4y^2 + 4y - 1
(7)
6 x^2 + 7xy + 14x + 2y^2 + 9y + 4
(8)
12x^2 - 17xy - 2x + 6y^2 + 2y - 4
(9)
10x^2 + 3xy + 10x - 4y^2 + 11y - 6
(10)
20 x^2 + 7xy - 7x - 6y^2 + 12y - 6
解答1．
(1)
x^2 + 3xy + 4x + 2y^2 +7y + 3 \\
= x^2 + (3y + 4)x + (2y + 1)(y + 3) \\
= (x + 2y + 1)(x + y + 3)
(2)
x^2 - 7xy + 5x + 12y^2 -17y + 6 \\
= x^2 + (-7y + 5)x + (4y - 3)(3y - 2) \\
= \{x - (4y - 3)\} \{x - (3y - 2) \} \\
= (x - 4y + 3)(x - 3y + 2)
(3)
x^2 + 2xy + 3x - 3y^2 + y + 2 \\
= x^2 + (2y + 3)x - (3y + 2)(y - 1) \\
= \{x + (3y + 2)\} \{x - (y - 1) \} \\
= (x + 3y + 2)(x - y + 1)
(4)
x^2 - 3xy - 3x + 2y^2 + 5y + 2 \\
= x^2 + (-3y - 3)x + (2y + 1)(y + 2) \\
= \{x - (2y + 1)\} \{x - (y + 2) \} \\
= (x - 2y - 1)(x - y - 2)
(5)
2x^2 - 5xy + 8x + 2y^2 - 7y + 6 \\
= 2x^2 + (-5y + 8)x + (y - 2)(2y - 3) \\
= \{ 2x - (y - 2)\} \{ x - (2y - 3) \} \\
= (2x - y + 2)(x - 2y + 3)
(6)
2x^2 - 2xy + x - 4y^2 + 4y - 1 \\
= 2x^2 + (-2y + 1)x - (2y - 1)^2 \\
= \{2x + (2y - 1)\} \{x - (2y - 1) \} \\
= (2x + 2y - 1)(x - 2y + 1)
(7)
6 x^2 + 7xy + 14x + 2y^2 + 9y + 4 \\
= 6x^2 + (7y + 14)x + (2y + 1)(y + 4) \\
= (3x + 2y + 1)(2x + y + 4)
(8)
12x^2 - 17xy - 2x + 6y^2 + 2y - 4 \\
= 12x^2 - 17xy - 2x + 2( 3y^2 + y - 2) \\
= 12x^2 + (-17y + 2)x + 2(3y - 2)(y + 1) \\
= \{ 4x - (3y - 2)\} \{ 3x - 2(y + 1) \} \\
= (4x - 3y + 2)(3x - 2y - 2)
(9)
10x^2 + 3xy + 10x - 4y^2 + 11y - 6 \\
= 10x^2 + 3xy + 10x + - ( 4y^2 - 11y + 6) \\
= 10x^2 + (3y + 10)x - (4y - 3)(y - 2) \\
= \{5x + (4y - 3)\} \{2x - (y - 2) \} \\
= (5x + 4y - 3)(2x - y + 2)
(10)
20 x^2 + 7xy - 7x - 6y^2 + 12y - 6 \\
= 20x^2 + (7y - 7)x - 6 (y - 1)^2 \\
= \{5x - 2 (y - 1)\} \{4x + 3 (y - 1) \} \\
= (5x - 2y + 2)(4x + 3y - 3)
演習問題2．
次の式を因数分解せよ．
(1)
(x^2 + 3x)^2 + 4(x^2 + 3x) + 3
(2)
(x^2 + 7x)^2 + 16(x^2 + 7x) + 60
(3)
x^4 - y^4
(4)
x^4 + 4
解答2．
(1)
(x^2 + 3x)^2 + 4(x^2 + 3x) + 3 \\
= \{ (x^2 + 3x) + 3 \} \{ (x^2 + 3x) + 1 \} \\
= (x^2 + 3x + 3)(x^2 + 3x + 1)
(2)
(x^2 + 7x)^2 + 16(x^2 + 7x) + 60 \\
= (x^2 + 7x + 10)(x^2 + 7x + 6) \\
= (x + 5)(x + 2)(x + 6)(x + 1) \\
= (x + 6)(x + 5)(x + 2)(x + 1)
(3)
x^4 - y^4 \\
= (x^2 + y^2)(x^2 - y^2) \\
= (x^2 + y^2)(x + y)(x - y)
(4)
x^4 + 4 \\
= x^4 + 4x^2 + 4 - 4x^2 \\
= (x^2 + 2)^2 - (2x)^2 \\
= (x^2 + 2x + 2)(x^2 - 2x + 2)
《PR》
受験生の2人に1人が利用する圧倒的なわかりやすさ！まずは無料でお試し。
関連ページ
解説ページ
次の解説ページ