$A = x^2 - 4x + 3$ および $B = -2x^2 + 2x + 1$ が与えられたとき、次の式を計算します。 (1) $3A + B$ (2) $2A - 3B$ (3) $A + 2(2A - B)$ (4) $-5A + 3B - 2(-3A + 4B)$

1. 問題の内容

$A = x^2 - 4x + 3$ および $B = -2x^2 + 2x + 1$ が与えられたとき、次の式を計算します。

(1) $3A + B$

(2) $2A - 3B$

(3) $A + 2(2A - B)$

(4) $-5A + 3B - 2(-3A + 4B)$

2. 解き方の手順

(1) $3A + B$

$3A = 3(x^2 - 4x + 3) = 3x^2 - 12x + 9$

$3A + B = (3x^2 - 12x + 9) + (-2x^2 + 2x + 1) = (3x^2 - 2x^2) + (-12x + 2x) + (9 + 1) = x^2 - 10x + 10$

(2) $2A - 3B$

$2A = 2(x^2 - 4x + 3) = 2x^2 - 8x + 6$

$3B = 3(-2x^2 + 2x + 1) = -6x^2 + 6x + 3$

$2A - 3B = (2x^2 - 8x + 6) - (-6x^2 + 6x + 3) = 2x^2 - 8x + 6 + 6x^2 - 6x - 3 = (2x^2 + 6x^2) + (-8x - 6x) + (6 - 3) = 8x^2 - 14x + 3$

(3) $A + 2(2A - B)$

$2A = 2(x^2 - 4x + 3) = 2x^2 - 8x + 6$

$2A - B = (2x^2 - 8x + 6) - (-2x^2 + 2x + 1) = 2x^2 - 8x + 6 + 2x^2 - 2x - 1 = 4x^2 - 10x + 5$

$2(2A - B) = 2(4x^2 - 10x + 5) = 8x^2 - 20x + 10$

$A + 2(2A - B) = (x^2 - 4x + 3) + (8x^2 - 20x + 10) = (x^2 + 8x^2) + (-4x - 20x) + (3 + 10) = 9x^2 - 24x + 13$

(4) $-5A + 3B - 2(-3A + 4B)$

$-5A = -5(x^2 - 4x + 3) = -5x^2 + 20x - 15$

$3B = 3(-2x^2 + 2x + 1) = -6x^2 + 6x + 3$

$-3A = -3(x^2 - 4x + 3) = -3x^2 + 12x - 9$

$4B = 4(-2x^2 + 2x + 1) = -8x^2 + 8x + 4$

$-2(-3A + 4B) = -2(-3x^2 + 12x - 9 - 8x^2 + 8x + 4) = -2(-11x^2 + 20x - 5) = 22x^2 - 40x + 10$

$-5A + 3B - 2(-3A + 4B) = (-5x^2 + 20x - 15) + (-6x^2 + 6x + 3) + (22x^2 - 40x + 10) = (-5x^2 - 6x^2 + 22x^2) + (20x + 6x - 40x) + (-15 + 3 + 10) = 11x^2 - 14x - 2$

3. 最終的な答え

(1) $x^2 - 10x + 10$

(2) $8x^2 - 14x + 3$

(3) $9x^2 - 24x + 13$

(4) $11x^2 - 14x - 2$