[演習問題] 1-2. 整式の掛け算と展開公式

解答1．
A - 2B \\ = 5x^4 - 2x^3 - 3x^2 - 4x + 2y - 9 -2(x^4 - x^3 - 2x^2 + x + 3y) \\ = 5x^4 - 2x^3 - 3x^2 - 4x + 2y - 9 -2x^4 + 2x^3 + 4x^2 - 2x - 6y \\ = 3x^4 + x^2 - 6x - 4y - 9

途中の式変形が分からない人は，分配法則を勉強し直しましょう．

解答2．
(1)
-2a(a - b - c) \\ = -2a \times a + (-2a) \times (-b) + (-2a) \times (-c) \\ = -2a^2 + 2ab + 2ac

(2)
3x(x -2y + z) \\ =3x \times x + 3x \times (-2y) + 3x \times z \\ = 3x^2 - 6xy + 3xz

(3)
-2a^2b(3a^2 - ab + 7b^2) \\ = -2a^2b \times 3a^2 + (-2a^2b) \times (- ab) + (-2a^2b) \times 7b^2 \\ = -6a^4b + 2a^3b^2 - 14a^2b^3

(4)
12xy(2x^2 + 12xy -3y^2) \\ = 12xy \times 2x^2 + (12xy)^2 + 12xy \times (-3y^2) \\ = 24x^3y + 144x^2y^2 - 36xy^3

(5)
11abc(2a - 11b - 3c - 4d) \\ = 11abc \times 2a + 11abc \times (- 11b) + 11abc \times (-3c) + 11abc \times (-4d) \\ = 22a^2bc - 121ab^2c - 33abc^2 - 44abcd

(6)
(a + b)(a + c + d) \\ = a(a + c + d) + b(a + c + d) \\ = a^2 +ac + ad +ab + bc + bd

解答3．

(1) (2a + b)^2 = 4a^2 + 4ab + b^2
(2) (3s + 2t)^2 = 9s^2 + 12st + 4t^2
(3) (12x + 7y)^2 = 144x^2 + 168xy + 49y^2
(4) (x + 5y)^2 = x^2 + 10xy + 25y^2
(5) (a - 2b)^2 = a^2 - 4ab + 4b^2
(6) (2a - 3b)^2 = 4a^2 - 12ab + 9b^2
(7) (11x - 7y)^2 = 121x^2 - 154xy + 49y^2
(8) (5x - 3y)^2 = 25x^2 - 30xy +9y^2
(9) (a + 2b)(a - 2b) = a^2 - 4b^2
(10) (4a + 3b)(4a - 3b) = 16a^2 - 9b^2
(11) (7x + 5y)(7x - 5y) = 49x^2 - 25y^2
(12) (3x - 8y)(3x + 8y) = 9x^2 - 64y^2
(13) (x + 2)(x - 1) = x^2 + x - 2
(14) (x + 4y)(x - 2y) = x^2 + 2xy - 8y^2
(15) (x + 5)(x + 3) = x^2 + 8x + 15
(16) (2x - 3)(3x + 4) = 6x^2 - x - 12
(17) (x + 3)(x - 3) = x^2 - 9
(18) (11a + 5b)^2 = 121a^2 + 110ab + 25b^2
(19) (2a - 3b)(x + 2y) = 2ax + 2ay - 3bx - 6by
(20) (a - 9b)^2 = a^2 - 18ab + 81b^2

(19)は引っかけです．必ずしも展開公式が使えるとは限りません．

解答4．

(1) (2a + b + c)^2 = 4a^2 + b^2 + c^2 + 4ab + 2bc + 4ca
(2) (a + 3b + 2c)^2 = a^2 + 9b^2 + 4c^2 + 6ab + 12bc + 4ca
(3) (2x - 3y + z)^2 = 4x^2 + 9y^2 + z^2 - 12xy - 6yz + 4zx
(4) (x - 5y - 6z)^2 = x^2 + 25y^2 + 36z^2 - 10xy + 60yz - 12zx
(5) (3x + 7y - 4z)^2 = 9x^2 + 49y^2 + 16z^2 + 42xy - 56yz - 24zx

解答5．
(1) (2a + b)^3 = (2a)^3 +3(2a)^2b + 3(2a)b^2 + b^3 = 8a^3 + 12a^2b + 6ab^2 + b^3
(2) (3s + 2t)^3 = 27s^3 + 54s^2t + 36st^2 + 8t^3
(3) (2x + 3y)^3 = 8x^3 + 36 x^2y + 54xy^2 + 27y^3
(4) (4x + 5y)^3 = 64x^3 + 240x^2y + 300xy^2 + 125y^3
(5) (a - 2b)^3 = a^3 - 6a^2b + 12ab^2 - 8b^3
(6) (5x - 3y)^3 = 125x^3 - 225x^2y + 135xy^2 -27y^3
(7) (2x - 7)^3 = 8x^3 - 84x^2 + 294x - 343
(8) (x - 3)^3 = x^3 - 9x^2 + 27x + 27
(9) (a - 2b)(a^2 + 2ab + 4b^2) = a^3 - 8b^3
(10) (4a + 3b)(16a^2 -12ab + 9b^2) = 64a^3 + 27b^3
(11) (x + 5y)(x^2 -5xy + 25y^2) = x^3 + 125y^3
(12) (3x + 2)(9x^2 - 6x + 4) = 27x^3 + 8
(13) (x - 2)(x^2 + 2x + 4) = x^3 - 8
(14) (x - 4y)(x^2 + 4xy + 16y^2) = x^3 - 64y^3
(15) (2a + b)(4a^2 - 2ab + b^2) = 8a^3 + b^3
(16) (2a - 3)(4a^2 + 6a + 9) = 8a^3 - 27
(17) (x + 3)^3 = x^3 + 9x^2 + 27x + 27
(18) (11a + 5b)^3 = 1331a^3 + 1815a^2b + 825ab^2 + 125b^3
(19) (a + b + 1)(a^2 + b^2 - ab - a - b + 1) = a^3 + b^3 -3ab + 1
(20) (2a - 7b)^2 = 4a^2 - 28ab + 49b^2

(20)は2乗であることを見落とさないようにしましょう．

解答6．
(1) (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4
(2) (x^2 - 2y^2)^2 = x^4 - 4x^2y^2 + 4y^4
(3) (x^2 + 2y^2)(x^2 - 2y^2) = x^4 - 4y^4
(4) (2x^2 + y^2)^3 = 8x^6 + 12x^4y^2 + 6x^2y^4 + y^6
(5) (a - 2b + c)(a + 2b + c) = \{ (a + c) - 2b \} \{ (a + c) + 2b \} = (a + c)^2 - (2b)^2 = a^2 - 4b^2 + c^2 + 2ca
(6) (2x - 3y + z)(2x - y + z) = \{ (2x + z) - 3y \} \{ (2x + z) - y \} = (2x + z)^2 - 4y(2x + z) + 3y^2 = 4x^2 + 4zx + z^2 - 8xy - 4yz + 3y^2 = 4x^2 + 3y^2 + z^2 - 8xy - 4yz + 4zx
(7) (x^2 - 5)^3 = x^6 - 15x^4 + 75x^2 - 125
(8) (x^3 - 3)^3 = x^9 - 9x^6 + 27x^3 - 27
(9) (a^2 - 2b^2)(a^4 + 2a^2b^2 + 4b^4) a^6 - 8b^6
(10) (a + b^2)^2 = a^2 + 2ab^2 + b^4
(11) (a + b + c + d)(a + b - c - d) = \{ (a + b) + (c + d) \} \{ (a + b) - (c + d) \} = (a + b)^2 - (c + d)^2 = a^2 + b^2 - c^2 - d^2 + 2ab - 2cd

(5) は (a + c) をひとつのまとまりとして見ましょう．
(6) は (2x + z) をひとつのまとまりとして見ましょう．
(11) は (a + b) と (c + d) をかたまりとして見ると，(a + b)(a - b) = a^2 - b^2 の公式が使えます．