多項式 $A$, $B$, $C$ が与えられたとき、$A - 2(B + 2C)$ を計算する。 $A = a^2 + 3ab - 2b^2$ $B = -2a^2 + 3ab - 5b^2$ $C = -5a^2 - 4ab + b^2$代数学多項式式の計算展開2025/7/91. 問題の内容多項式 AAA, BBB, CCC が与えられたとき、A−2(B+2C)A - 2(B + 2C)A−2(B+2C) を計算する。A=a2+3ab−2b2A = a^2 + 3ab - 2b^2A=a2+3ab−2b2B=−2a2+3ab−5b2B = -2a^2 + 3ab - 5b^2B=−2a2+3ab−5b2C=−5a2−4ab+b2C = -5a^2 - 4ab + b^2C=−5a2−4ab+b22. 解き方の手順まず、B+2CB + 2CB+2C を計算する。B+2C=(−2a2+3ab−5b2)+2(−5a2−4ab+b2)B + 2C = (-2a^2 + 3ab - 5b^2) + 2(-5a^2 - 4ab + b^2)B+2C=(−2a2+3ab−5b2)+2(−5a2−4ab+b2)B+2C=−2a2+3ab−5b2−10a2−8ab+2b2B + 2C = -2a^2 + 3ab - 5b^2 - 10a^2 - 8ab + 2b^2B+2C=−2a2+3ab−5b2−10a2−8ab+2b2B+2C=(−2−10)a2+(3−8)ab+(−5+2)b2B + 2C = (-2 - 10)a^2 + (3 - 8)ab + (-5 + 2)b^2B+2C=(−2−10)a2+(3−8)ab+(−5+2)b2B+2C=−12a2−5ab−3b2B + 2C = -12a^2 - 5ab - 3b^2B+2C=−12a2−5ab−3b2次に、2(B+2C)2(B + 2C)2(B+2C) を計算する。2(B+2C)=2(−12a2−5ab−3b2)2(B + 2C) = 2(-12a^2 - 5ab - 3b^2)2(B+2C)=2(−12a2−5ab−3b2)2(B+2C)=−24a2−10ab−6b22(B + 2C) = -24a^2 - 10ab - 6b^22(B+2C)=−24a2−10ab−6b2最後に、A−2(B+2C)A - 2(B + 2C)A−2(B+2C) を計算する。A−2(B+2C)=(a2+3ab−2b2)−(−24a2−10ab−6b2)A - 2(B + 2C) = (a^2 + 3ab - 2b^2) - (-24a^2 - 10ab - 6b^2)A−2(B+2C)=(a2+3ab−2b2)−(−24a2−10ab−6b2)A−2(B+2C)=a2+3ab−2b2+24a2+10ab+6b2A - 2(B + 2C) = a^2 + 3ab - 2b^2 + 24a^2 + 10ab + 6b^2A−2(B+2C)=a2+3ab−2b2+24a2+10ab+6b2A−2(B+2C)=(1+24)a2+(3+10)ab+(−2+6)b2A - 2(B + 2C) = (1 + 24)a^2 + (3 + 10)ab + (-2 + 6)b^2A−2(B+2C)=(1+24)a2+(3+10)ab+(−2+6)b2A−2(B+2C)=25a2+13ab+4b2A - 2(B + 2C) = 25a^2 + 13ab + 4b^2A−2(B+2C)=25a2+13ab+4b23. 最終的な答え25a2+13ab+4b225a^2 + 13ab + 4b^225a2+13ab+4b2