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Descompuneti in factori :
a) x⁴ + 8x²+ 16
b) 9y - 6x³y + x la a 6
Va roggg

Răspuns :

[tex]a)~~x^4 + 8x^2+ 16 = \Big(x^2\Big)^2 + 2\cdot x^2 \cdot 4 + 4^2 =\boxed{\Big(x^2 + 4\Big)^2} \\ \\ \\ \\ b) ~~9y^2 - 6x^3y + x^6 = \Big(3y\Big)^2 + 2\cdot 3y \cdot (-x^3) + \Big(-x^3\Big)^2 = \boxed{\Big(3y-x^3\Big)^2}[/tex]



Icecon2005
a) x⁴ + 8x²+ 16 =[tex](x^{2} +4)^{2}=( x^{2} +2)( x^{2} +2) [/tex]

b) [tex]9y^{2} -6x^{3}y+x^{6}=(3y-x^{3})^{2}=(3y-x^{3})( 3y-x^{3})[/tex]

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