Answer:
The given quadratic equation \(4n^2+28n+49\) factored as \((2n+7)^2=(2n+7)(2n+7)\)
Step-by-step explanation:
Given quadratic equation \(4n^2+28n+49\)
We have to factorize the given quadratic equation and make the grid of factors.
Consider the given quadratic equation \(4n^2+28n+49\)
Using algebraic identity,
\((a+b)^2=a^2+2ab+b^2\)
On comparing, we get,
\(a^2=4n^2\\\\ \Rightarrow a=2n\)
also, \(b^2=49\\\\\Rightarrow b=7\)
Thus, the given quadratic equation \(4n^2+28n+49\) factored as \((2n+7)^2=(2n+7)(2n+7)\)
The grid is shown in attachment below.
