Question

Valendo 20 pontos fatore:

a) 6x² + 12xy + 6y²
b) 4x² - 16y²
c) 3x + 3y + 6x² - 6y²

Answer 1

Oie, Td Bom?!

a)

$= 6x {}^{2} + 12xy + 6y {}^{2}$

$= 6(x {}^{2} + 2xy + y {}^{2} )$

$= 6(x + y) {}^{2}$

b)

$= 4x {}^{2} - 16y {}^{2}$

$= 4(x {}^{2} - 4 y{}^{2} )$

$= 4 \: . \: x {}^{2} - 2 {}^{2} y {}^{2}$

$= 4 \: . \: x {}^{2} - (2y) {}^{2}$

$= 4(x - 2y) \: . \: (x + 2y)$

c)

$= 3x + 3y + 6x {}^{2} - 6y {}^{2}$

$= 3(x + y + 2x {}^{2} - 2y {}^{2} )$

$= 3(x + y + 2(x {}^{2} - y {}^{2} ))$

$= 3(x + y + 2(x - y) \: . \: (x + y))$

$= 3(x + y) \: . \: (1 + 2(x - y))$

$= 3(x + y) \: . \: (1 + 2x - 2y)$

Att. Makaveli1996

Answer 2

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$\tt a)~\sf 6x^2+12xy+6y^2=6\cdot(x^2+2xy+y^2)\\\implies\huge\boxed{\boxed{\boxed{\boxed{\sf6\cdot(x+y)^2}}}}\\\tt b)~\sf4x^2-16y^2=\boxed{\boxed{\boxed{\boxed{\sf (2x+4y)\cdot(2x-4y)}}}}\\\tt c)~\sf3x+3y+6x^2-6y^2\\\sf3\cdot(x+y)+6\cdot(x^2-y^2)\\\sf3\cdot(x+y)+6\cdot(x+y)(x-y)\\\huge\boxed{\boxed{\boxed{\boxed{\sf3\cdot(x+y)\left[1+2\cdot(x-y)\right]}}}}$