Question

Factor completely 4n 2 + 28n + 49

Answer 1

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

Answer 2

Factor of Equation
`4n^2+28n+49` is
`\left(2n+7\right)\left(2n+7\right)`

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)`