Answer:
Factor of Equation \(4n^2+28n+49\) is \(\left(2n+7\right)\left(2n+7\right)\)
Step-by-step explanation:
Given : Equation \(4n^2+28n+49\)
We have to factorize the given equation completely.
Consider the given equation \(4n^2+28n+49\)
We can rewrite \(4n^2=2^2n^2\) , we get,
\(=2^2n^2+28n+49\)
We can rewrite \(49=7^2\) , we get,
\(=2^2n^2+28n+7^2\)
Apply exponent rule, \(a^mb^m=\left(ab\right)^m\)
\(=\left(2n\right)^2+28n+7^2\)
Apply identity, \(\left(a+b\right)^2=a^2+2ab+b^2\) ,we get,
\(\left(2n+7\right)^2\)
Thus, Factor of Equation \(4n^2+28n+49\) is \(\left(2n+7\right)\left(2n+7\right)\)
