Question

Разложи на множители многчлен

Answer 1

$\displaystyle\bf\\y^{4} +13y^{3} +56y^{2} +92y+48=y^{4} +12y^{3} +44y^{2} +48y+y^{3} +12y^{2} +44y+48=\\\\=(y^{4} +12y^{3} +44y^{2} +48y)+(y^{3} +12y^{2} +44y+48)=\\\\=y\cdot(y^{3} +12y^{2}+44y+48)+( y^{3} +12y^{2}+44y+48)=\\\\=(y^{3} +12y^{2}+44y+48)\cdot(y+1)=(y+1)\cdot(y^{3} +2y^{2}+10y^{2} +20y+24y+48)=\\\\=(y+1)\cdot\Big[(y^{3} +2y^{2})+(10y^{2} +20y)+(24y+48)\Big]=$

$\displaystyle\bf\\=(y+1)\Big[y^{2} (y+2)+10y(y+2)+24(y+2)\Big]=\\\\=(y+1)(y+2)(y^{2} +10y+24)=(y+1)(y+2)(y+4)(y+6)$