81-49n^4 difference of two squares?
2010-03-25 3:04 pm
Determine whether 81-49n^4 is a difference of two squares. If so factor it. If not explain why.
回答 (7)
2010-03-25 3:21 pm
✔ 最佳答案
Hi,Short Answer: (9-7n)(9+7n)
Comments:
As others have said, this is the difference of two squares. Here's a quick way to factor it:
Take the square root of each term and write these roots in two sets of parentheses like this:
(9 7n)(9 7n)
Now, put a + sign in one binomial and a negative in the other. It doesn't matter whether the positive sign goes in the first or second binomial. So, you get this either this:
(9-7n)(9+7n) or this: (9+7n)(9-7n). They are both equally correct.
Hope this helps.
FE
2010-03-25 3:11 pm
81 – 49n⁴ = (9)² – (7n²)²
.............. = (9 + 7n²)(9 – 7n²)
.............. = (9 + 7n²)(9 – 7n²)
2010-03-25 3:30 pm
a^2 - b^2 ≡ (a + b)(a - b)
81 - 49n^4
= 9^2 - (7n^2)^2
= (9 + 7n^2)(9 - 7n^2)
81 - 49n^4
= 9^2 - (7n^2)^2
= (9 + 7n^2)(9 - 7n^2)
2010-03-25 3:23 pm
( 9 - 7 n ² ) ( 9 + 7 n ² )
2010-03-25 3:10 pm
Yes it is
-(7n^2-9) (7n^2+9)
-(7n^2-9) (7n^2+9)
2010-03-25 3:09 pm
this is an example of a difference of two squares.
factor this out to verify to yourself that this equation is indeed a difference of two squares
(81-49n^4) =
(9-7n^2)(9+7n^2) ---> proof that this is an example of the difference of two squares
factor this out to verify to yourself that this equation is indeed a difference of two squares
(81-49n^4) =
(9-7n^2)(9+7n^2) ---> proof that this is an example of the difference of two squares
收錄日期: 2021-05-01 13:07:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100325070451AAsfpog