Question

Factor completely 4n 2 + 28n + 49

Answer 1

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 $\left(2n+7\right)\left(2n+7\right)$

Answer 2

4n^2 + 28n + 49 = 0 factorises to:
(2n+7)(2n+7) = 0