Again Factoring help ?
回答 (7)
2009-05-02 9:21 am
✔ 最佳答案
Hi,Problem : p^2 +p - (a + 1)(a +2)
Now LOOK (a + 2) - (a + 1) = a+ 2 - a - 1 = 1
Hence, p^2 + {(a + 2) - (a + 1)}p - (a + 2)(a + 1)
= p^2 + (a + 2)p - (a + 1)p - (a + 2)(a + 1)
= p{p + a + 2} - (a + 1) {p + a + 2}
= (p + a +2)(p - a - 1) <==FACTORED
Hope i helped u :)
2009-05-02 3:46 pm
Dhonilover, check your wording again!!!
(p-a-1)(p+a+2)
(p-a-1)(p+a+2)
2009-05-02 6:30 pm
Answer:
-(a - p + 1)(a + p + 2)
-(a - p + 1)(a + p + 2)
2009-05-02 4:33 pm
p² + p - (a + 1)(a + 2)
= p² + p - (a² + 3a + 2)
= p² + p - a² - 3a - 2
= p² - a² + p- 3a - 2
=p² - a² - 2a -1 + p-a- 1
=p² - (a² + 2a +1 )+(p-a- 1)
=p² - (a + 1)²+(p-a- 1)
=(p + a + 1)(p - a - 1)+(p-a- 1)
=(p - a - 1)(p + a + 1+1)
=(p - a - 1)(p + a + 2)
= p² + p - (a² + 3a + 2)
= p² + p - a² - 3a - 2
= p² - a² + p- 3a - 2
=p² - a² - 2a -1 + p-a- 1
=p² - (a² + 2a +1 )+(p-a- 1)
=p² - (a + 1)²+(p-a- 1)
=(p + a + 1)(p - a - 1)+(p-a- 1)
=(p - a - 1)(p + a + 1+1)
=(p - a - 1)(p + a + 2)
2009-05-02 4:14 pm
p^2 + p - (a + 1)(a + 2)
= p^2 + p - (a^2 + a + 2a + 2)
= p^2 + p - a^2 - 3a - 2
= p^2 + ap + 2p - ap - a^2 - 2a - p - a - 2
= (p^2 + ap + 2p) - (ap + a^2 + 2a) - (p + a + 2)
= p(p + a + 2) - a(p + a + 2) - 1(p + a + 2)
= (p + a + 2)(p - a - 1)
= p^2 + p - (a^2 + a + 2a + 2)
= p^2 + p - a^2 - 3a - 2
= p^2 + ap + 2p - ap - a^2 - 2a - p - a - 2
= (p^2 + ap + 2p) - (ap + a^2 + 2a) - (p + a + 2)
= p(p + a + 2) - a(p + a + 2) - 1(p + a + 2)
= (p + a + 2)(p - a - 1)
2009-05-02 3:44 pm
= p(p+p)-(a+1)(a+2)
= (p-a-1)(p+a+2)
= (p-a-1)(p+a+2)
2009-05-02 3:48 pm
(p+a+2)(p-a+1)
收錄日期: 2021-05-01 12:21:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090502004049AANb1gc