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

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

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factor 4u^2+28u+49

factor

4 u^{2} + 28 u + 49

(2 u + 7)^{2}

Solution steps

4 u^{2} + 28 u + 49

Rewrite in the form of

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

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

49 = 7^{2}

28 u = 2 \cdot 2 u \cdot 7

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

Apply Perfect Square Formula:

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

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

= (2 u + 7)^{2}

2 (h + 2) + 7 = - 3

0.04 m^{2} - n^{2}

(\frac{2}{9})^{- 1}

6 t (t + 4) = t - 20

25 6^{\frac{5}{4}}