基礎数学 No.3 2004. 10. 2
1.3 式の展開と因数分解（解答）
担当：市原問題 7 展開公式を利用して, 次の値を求めなさい. (1) 7×1997 = 7×(2000−3) = 14000−21 = 13979 (2) 1022 = (100 + 2)2 = 10000 + 400 + 4 = 10404 (3) 392 = (40−1)2 = 1600−80 + 1 = 1521
(4) 103×97 = (100 + 3)(100−3) = 10000−9 = 9991
(5) 104×99 = (100 + 4)(100−1) = 10000 + 300−4 = 10296 (6) 1023 = (100 + 2)3 = 1000000 + 60000 + 1200 + 8 = 1061208 (7) 993 = (100−1)3 = 1000000−30000 + 300−1 = 970299
問題 8 次の各式を因数分解しなさい．
(1) x2−3x−40 = (x−8)(x+ 5) (2) z8−5z4+ 4 = (z4−4)(z4−1)
= (z2+ 2)(z2 −2)(z2+ 1)(z2−1) = (z2+ 2)(z2−2)(z2+ 1)(z+ 1)(z−1) (3) 6x2+ 5x+ 1 = (3x+ 1)(2x+ 1)
(4) 8x3+ 27 = (2x)3+ 33 = (2x+ 3)((2x)2+ 6x+ 9) = (2x+ 3)(4x2+ 6x+ 9) (5) 6t2+ 10t+ 4 = 2(3t2+ 5t+ 2) = 2(3t+ 2)(t+ 1)
(6) 3A3−7A2+ 2A=A(3A2−7A+ 2) =A(3A−1)(A−2)
問題 9 次のxに関する式をa(x−b)2+cの形に表しなさい．
(1) 3(x2+ 6x+ 9)−1 = 3(x+ 3)2−1
(2) 2x2−2x+ 1 = 2 µ
x2−x+ 1 2
¶
= 2 õ
x− 1 2
¶2
−1 4 +1
2
!
= 2 õ
x− 1 2
¶2 +1
4
!
= 2 µ
x−1 2
¶2 + 1
2
(3) 4x2+ 6x−1 = 4 µ
x2+ 3 2x− 1
4
¶
= 4 õ
x+ 3 4
¶2
− 9 16 − 1
4
!
= 4 õ
x+ 3 4
¶2
−13 16
!
= 4 µ
x+ 3 4
¶2
−13 4
