標題:

F.2 Maths Factorization

發問:

1. m^ 4- 18m^ 2+ 81 2. 16p^ 4- 8p^ 2q^2+ q^4

免費註冊體驗

 

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

1. m^ 4- 18m^ 2+ 81 [m^2]^2 - 18(m^ 2) + 81 (m^2 - 9)((m^2 - 9) (m +3)(m - 3)(m +3)(m - 3) (m +3)^2 (m - 3)^2 2. 16p^ 4- 8p^ 2q^2+ q^4 16(p^ 2)^2 - 8(pq)^2+ (q^2)^2 (4p^2 - q^2)(4p^2 - q^2) ((2p)^2 - q^2)((2p)^2 - q^2) (2p +q)(2p - q)(2p +q)(2p - q) (2p +q)^2 (2p - q)^2

其他解答:

m^4 = (m^2)^2 m^ 4- 18m^ 2+ 81 = (m^2)^2 - 2.9.m^2 + 9^2 = (m^2 - 9)^2 or let A = m^2 then m^ 4- 18m^ 2+ 81 = A^2 - 2.9.A + 9^2 = (A - 9)^2 = (m^2 - 9)^2 Similar 16p^ 4- 8p^2q^2+ q^4 = (4p^ 2)^2 - 2.4p^ 2.q^2 + (q^2)^2 = (4p^ 2 - q^2)^2 2013-09-06 13:22:31 補充: a^2 - b^2 = (a+b)(a-b) m^2 - 9 = (m^2 - 3^2) = (m+3)(m-3) => m^ 4- 18m^ 2+ 81 = (m^2 - 9)^2 = [(m+3)(m-3)]^2 = (m+3)^2(m-3)^2 4p^ 2 - q^2 = (2p)^2 - q^2 = (2p + q)(2p - q) => 16p^ 4- 8p^2q^2+ q^4 = (2p + q)^2.(2p - q)^2 2013-09-07 02:09:18 補充: https://www.dropbox.com/s/1bpt9fyqj2kp5sn/20130906PIC02.png
arrow
arrow

    reyesb28mi45 發表在 痞客邦 留言(0) 人氣()