What is factor 4n^2+28n+49 ?

The solution to factor 4n^2+28n+49 is (2n+7)^2

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factor 4n^2+28n+49

factor

4 n^{2} + 28 n + 49

(2 n + 7)^{2}

Solution steps

4 n^{2} + 28 n + 49

Rewrite in the form of

a^{2} + 2 ab + b^{2} :

4 n^{2} = (2 n)^{2}

49 = 7^{2}

28 n = 2 \cdot 2 n \cdot 7

= (2 n)^{2} + 2 \cdot 2 n \cdot 7 + 7^{2}

Apply Perfect Square Formula:

a^{2} + 2 ab + b^{2} = (a + b)^{2}

(2 n)^{2} + 2 \cdot 2 n \cdot 7 + 7^{2} = (2 n + 7)^{2}

= (2 n + 7)^{2}

14 4^{- \frac{1}{2}}

x^{4} + x^{5} y^{4}

\frac{1}{144}

x^{3} - 14 x^{2} + 49 x

- 2 x^{2} + 4 x - 29 = 0