1(a). Factorize 8x^3-27 = (2x-3)(4x^2+6x+9) I know how to do 1(b). Hence factorize 8x^3-8x-15 = (2x-3)(4x^2+6x+5) I want the steps?
回答 (1)
2015-10-23 3:11 pm
1.
a^3-b^3=(a-b)(a^2+ab+b^2)
(2x)^3-3^3=(2x-3)((2x)^2+(2x)*3+3^2)=(2x-3)(4x^2+6x+9)
2.
8x^3-8x-15=8x^3-27-8x-15+27
=(2x)^3-3^3-(8x-12)
=(2x-3)(4x^2+6x+9)-4(2x-3)
=(2x-3)(4x^2+6x+9-4)
=(2x-3)(4x^2+6x+5)
a^3-b^3=(a-b)(a^2+ab+b^2)
(2x)^3-3^3=(2x-3)((2x)^2+(2x)*3+3^2)=(2x-3)(4x^2+6x+9)
2.
8x^3-8x-15=8x^3-27-8x-15+27
=(2x)^3-3^3-(8x-12)
=(2x-3)(4x^2+6x+9)-4(2x-3)
=(2x-3)(4x^2+6x+9-4)
=(2x-3)(4x^2+6x+5)
收錄日期: 2021-04-18 13:57:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151023065513AABuqrc