Question

Factor completely 4n^2+28n+49

Answer 1

(2n+7)²

Explanation:

Given expression is

4n²+28n+49

(a+b)² = a²+2ab+b² is formula to find factors of any quadratic expression

4n² is square of 2n.

49 is square of 7.

And 28n is 2(2n)(7).

a = 2n and b = 7

Hence, we can put above value in formula.

(2n+7)² = ( (2n)²+2(2n)(7) + (7)² )

(2n+7)² = ( 4n² + 28n +49)

Hence, (2n+7)² are factors of 4n²+28n+49.

Answer 2

2n+7)(2n+7)

Explanation:

We are given an expression 4n² + 28n + 49 and we have to factorize it

4n² + 28n + 49

We can write 4n² as (2n)²

We can write 49 as 7²

We can also write 28n as 2×7×2n

so rewriting the above expression

(2n)² + 2×7×2n + 7²

Apply perfect square formula:
`(a+b)^2=a^2+2ab+b^2`

a = 2n ; b = 7

the expression become

(2n+7)²

It can also be written as (2n+7)(2n+7)