$P = 2x^2 + x - 1$、 $Q = -4x^2 + 4x + 2$のとき、以下の計算をせよ。 (1) $P - Q$ (2) $5(P - 2Q) - (3P - 7Q)$代数学多項式式の計算展開2025/5/171. 問題の内容P=2x2+x−1P = 2x^2 + x - 1P=2x2+x−1、 Q=−4x2+4x+2Q = -4x^2 + 4x + 2Q=−4x2+4x+2のとき、以下の計算をせよ。(1) P−QP - QP−Q(2) 5(P−2Q)−(3P−7Q)5(P - 2Q) - (3P - 7Q)5(P−2Q)−(3P−7Q)2. 解き方の手順(1) P−QP - QP−Qの計算P−Q=(2x2+x−1)−(−4x2+4x+2)P - Q = (2x^2 + x - 1) - (-4x^2 + 4x + 2)P−Q=(2x2+x−1)−(−4x2+4x+2)P−Q=2x2+x−1+4x2−4x−2P - Q = 2x^2 + x - 1 + 4x^2 - 4x - 2P−Q=2x2+x−1+4x2−4x−2P−Q=(2x2+4x2)+(x−4x)+(−1−2)P - Q = (2x^2 + 4x^2) + (x - 4x) + (-1 - 2)P−Q=(2x2+4x2)+(x−4x)+(−1−2)P−Q=6x2−3x−3P - Q = 6x^2 - 3x - 3P−Q=6x2−3x−3(2) 5(P−2Q)−(3P−7Q)5(P - 2Q) - (3P - 7Q)5(P−2Q)−(3P−7Q)の計算まず、5(P−2Q)5(P - 2Q)5(P−2Q)と(3P−7Q)(3P - 7Q)(3P−7Q)を計算します。5(P−2Q)=5P−10Q5(P - 2Q) = 5P - 10Q5(P−2Q)=5P−10Q3P−7Q=3P−7Q3P - 7Q = 3P - 7Q3P−7Q=3P−7Q次に、5(P−2Q)−(3P−7Q)5(P - 2Q) - (3P - 7Q)5(P−2Q)−(3P−7Q)を計算します。5(P−2Q)−(3P−7Q)=(5P−10Q)−(3P−7Q)5(P - 2Q) - (3P - 7Q) = (5P - 10Q) - (3P - 7Q)5(P−2Q)−(3P−7Q)=(5P−10Q)−(3P−7Q)5(P−2Q)−(3P−7Q)=5P−10Q−3P+7Q5(P - 2Q) - (3P - 7Q) = 5P - 10Q - 3P + 7Q5(P−2Q)−(3P−7Q)=5P−10Q−3P+7Q5(P−2Q)−(3P−7Q)=(5P−3P)+(−10Q+7Q)5(P - 2Q) - (3P - 7Q) = (5P - 3P) + (-10Q + 7Q)5(P−2Q)−(3P−7Q)=(5P−3P)+(−10Q+7Q)5(P−2Q)−(3P−7Q)=2P−3Q5(P - 2Q) - (3P - 7Q) = 2P - 3Q5(P−2Q)−(3P−7Q)=2P−3QPPPとQQQを代入します。2P−3Q=2(2x2+x−1)−3(−4x2+4x+2)2P - 3Q = 2(2x^2 + x - 1) - 3(-4x^2 + 4x + 2)2P−3Q=2(2x2+x−1)−3(−4x2+4x+2)2P−3Q=4x2+2x−2+12x2−12x−62P - 3Q = 4x^2 + 2x - 2 + 12x^2 - 12x - 62P−3Q=4x2+2x−2+12x2−12x−62P−3Q=(4x2+12x2)+(2x−12x)+(−2−6)2P - 3Q = (4x^2 + 12x^2) + (2x - 12x) + (-2 - 6)2P−3Q=(4x2+12x2)+(2x−12x)+(−2−6)2P−3Q=16x2−10x−82P - 3Q = 16x^2 - 10x - 82P−3Q=16x2−10x−83. 最終的な答え(1) 6x2−3x−36x^2 - 3x - 36x2−3x−3(2) 16x2−10x−816x^2 - 10x - 816x2−10x−8