Question
$4a^2 + b^2 + 4ab + 8a + 4b + 4 =$ ?
✓
Answer
We know that,
$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
$2a^2 + b^2 + 4 + 4ab + 8a + 4b + 4$
$= 4a^2 + b^2 + 4 + 4ab + 8a + 4b$
$= (2a^2) + b^2 + 2^2 + 2(2a)b + 2(2a) (2) + 2(2b)$
$= (2a + b + 2)^2$
$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
$2a^2 + b^2 + 4 + 4ab + 8a + 4b + 4$
$= 4a^2 + b^2 + 4 + 4ab + 8a + 4b$
$= (2a^2) + b^2 + 2^2 + 2(2a)b + 2(2a) (2) + 2(2b)$
$= (2a + b + 2)^2$
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