What is factor 1+4y+4y^2 ?

The solution to factor 1+4y+4y^2 is (2y+1)^2

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factor 1+4y+4y^2

factor

1 + 4 y + 4 y^{2}

(2 y + 1)^{2}

Solution steps

1 + 4 y + 4 y^{2}

Write in the standard form

= 4 y^{2} + 4 y + 1

Rewrite in the form of

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

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

1 = 1^{2}

4 y = 2 \cdot 2 y \cdot 1

= (2 y)^{2} + 2 \cdot 2 y \cdot 1 + 1^{2}

Apply Perfect Square Formula:

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

(2 y)^{2} + 2 \cdot 2 y \cdot 1 + 1^{2} = (2 y + 1)^{2}

= (2 y + 1)^{2}

\frac{24}{3}

4 x + 2 < - 6 or 2 x - 6 \geq - 14

6 x - 30 + 8 x + 28 = 180

1 3^{2}

\frac{2 x}{3} + 9 = \frac{1}{2}