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How do you factor 4n^2-49? | HIX Tutor

<mathjax>#4n^2-49 = (2n)^2-7^2 = (2n-7)(2n+7)#</mathjax>
...using the difference of squares identity:
<mathjax>#a^2-b^2 = (a-b)(a+b)#</mathjax>

<mathjax>#4n^2-49 = (2n)^2-7^2 = (2n-7)(2n+7)#
...using the difference of squares identity:
<mathjax>#a^2-b^2 = (a-b)(a+b)#

Ethan Adkins

To factor \(4n^2 - 49\), we recognize it as a difference of squares because it can be expressed as \( (2n)^2 - 7^2 \). 

The formula for factoring a difference of squares is \(a^2 - b^2 = (a + b)(a - b)\). 

So, we rewrite \(4n^2 - 49\) as \((2n + 7)(2n - 7)\). 

Therefore, the factored form of \(4n^2 - 49\) is \((2n + 7)(2n - 7)\).

How do you factor #4n^2-49#?