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a fost răspuns

Descompunetiin factori grupand mai intai convenabil termeni
5ax²+4ax³+10a²x+8a²x²=
12x²y³+4x²y+9y²z²+3z²=
5xy+2x²y+10x²y²+4x³y²=
2x²+3a²x-4ax-6a³=
va rog

Răspuns :

Renatemambouko
5ax²+4ax³+10a²x+8a²x²=
=5ax²+10a²x+4ax³+8a²x²=
=5ax(x+a)+4ax²(x+a)=
=(x+a)(5a+4ax²)=
=a(x+a)(5+4x²)=


 12x²y³+4x²y+9y²z²+3z²=
=12x²y³+9y²z²+4x²y+3z²=
=3y²(4x²y+3z²)+4x²y+3z²=
=(4x²y+3z²)(3y²+1)


5xy+2x²y+10x²y²+4x³y²=
=5xy+10x²y²+2x²y+4x³y²=
=5xy(1+2xy)+2x²y(1+2x²y)=
=(1+2xy)(5xy+2x²y)


2x²+3a²x-4ax-6a³=
=x(2x+3a²)-2a(2x+3a²)=
=(2x+3a²)(x-2a)
















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