Factors of $a^{2} + 4 a + 4$ are:

(a + 2)^{2}

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(a + 1)^{2}

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(a - 2)^{2}

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(a - 1)^{2}

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Solution

The correct option is A $(a + 2)^{2}$
$a^{2} + 4 a + 4 = a^{2} + 2 (a) (2) + (2)^{2}$
$= (a + 2)^{2}$

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, find the value of x.

$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . \infty = \frac{b}{6} e$ . Find $b$

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Factors of a2+4a+4 are:

The correct option is A (a+2)2a2+4a+4=a2+2(a)(2)+(2)2=(a+2)2

Factors of $a^{2} + 4 a + 4$ are:

The correct option is A $(a + 2)^{2}$
$a^{2} + 4 a + 4 = a^{2} + 2 (a) (2) + (2)^{2}$
$= (a + 2)^{2}$