$A = 2x^2 - 4x - 5$ と $B = 3x^2 - 2x + 2$ が与えられています。次の式を計算します。 (1) $A + B$ (2) $B - A$ (3) $-2A - 3B$ (4) $3(A+B) - 2(A-B)$代数学多項式式の計算展開2025/6/221. 問題の内容A=2x2−4x−5A = 2x^2 - 4x - 5A=2x2−4x−5 と B=3x2−2x+2B = 3x^2 - 2x + 2B=3x2−2x+2 が与えられています。次の式を計算します。(1) A+BA + BA+B(2) B−AB - AB−A(3) −2A−3B-2A - 3B−2A−3B(4) 3(A+B)−2(A−B)3(A+B) - 2(A-B)3(A+B)−2(A−B)2. 解き方の手順(1) A+B=(2x2−4x−5)+(3x2−2x+2)A + B = (2x^2 - 4x - 5) + (3x^2 - 2x + 2)A+B=(2x2−4x−5)+(3x2−2x+2)=2x2−4x−5+3x2−2x+2= 2x^2 - 4x - 5 + 3x^2 - 2x + 2=2x2−4x−5+3x2−2x+2=(2x2+3x2)+(−4x−2x)+(−5+2)= (2x^2 + 3x^2) + (-4x - 2x) + (-5 + 2)=(2x2+3x2)+(−4x−2x)+(−5+2)=5x2−6x−3= 5x^2 - 6x - 3=5x2−6x−3(2) B−A=(3x2−2x+2)−(2x2−4x−5)B - A = (3x^2 - 2x + 2) - (2x^2 - 4x - 5)B−A=(3x2−2x+2)−(2x2−4x−5)=3x2−2x+2−2x2+4x+5= 3x^2 - 2x + 2 - 2x^2 + 4x + 5=3x2−2x+2−2x2+4x+5=(3x2−2x2)+(−2x+4x)+(2+5)= (3x^2 - 2x^2) + (-2x + 4x) + (2 + 5)=(3x2−2x2)+(−2x+4x)+(2+5)=x2+2x+7= x^2 + 2x + 7=x2+2x+7(3) −2A−3B=−2(2x2−4x−5)−3(3x2−2x+2)-2A - 3B = -2(2x^2 - 4x - 5) - 3(3x^2 - 2x + 2)−2A−3B=−2(2x2−4x−5)−3(3x2−2x+2)=−4x2+8x+10−9x2+6x−6= -4x^2 + 8x + 10 - 9x^2 + 6x - 6=−4x2+8x+10−9x2+6x−6=(−4x2−9x2)+(8x+6x)+(10−6)= (-4x^2 - 9x^2) + (8x + 6x) + (10 - 6)=(−4x2−9x2)+(8x+6x)+(10−6)=−13x2+14x+4= -13x^2 + 14x + 4=−13x2+14x+4(4) 3(A+B)−2(A−B)=3A+3B−2A+2B3(A+B) - 2(A-B) = 3A + 3B - 2A + 2B3(A+B)−2(A−B)=3A+3B−2A+2B=(3A−2A)+(3B+2B)= (3A - 2A) + (3B + 2B)=(3A−2A)+(3B+2B)=A+5B= A + 5B=A+5B=(2x2−4x−5)+5(3x2−2x+2)= (2x^2 - 4x - 5) + 5(3x^2 - 2x + 2)=(2x2−4x−5)+5(3x2−2x+2)=2x2−4x−5+15x2−10x+10= 2x^2 - 4x - 5 + 15x^2 - 10x + 10=2x2−4x−5+15x2−10x+10=(2x2+15x2)+(−4x−10x)+(−5+10)= (2x^2 + 15x^2) + (-4x - 10x) + (-5 + 10)=(2x2+15x2)+(−4x−10x)+(−5+10)=17x2−14x+5= 17x^2 - 14x + 5=17x2−14x+53. 最終的な答え(1) 5x2−6x−35x^2 - 6x - 35x2−6x−3(2) x2+2x+7x^2 + 2x + 7x2+2x+7(3) −13x2+14x+4-13x^2 + 14x + 4−13x2+14x+4(4) 17x2−14x+517x^2 - 14x + 517x2−14x+5