Question

50БАЛОВ!!!!!
Разложите на множители способом группировки

Answer 1

$\displaystyle\bf\\1)\\\\xy+xz+y+z=(xy+y)+(xz+z)=\\\\=y\cdot(x+1)+z\cdot(x+1)=(x+1)\cdot(y+z)\\\\2)\\\\m+kn+n+km=(m+km)+(n+kn)=\\\\=m\cdot(1+k)+n\cdot(1+k)=(1+k)\cdot(m+n)\\\\3)\\\\m-kn+n-km=(m-km)+(n-kn)=\\\\=m\cdot(1-k)+n\cdot(1-k)=(1-k)\cdot(m+n)\\\\4)\\\\mn+5n+5k+km=(mn+km)+(5n+5k)=\\\\=m\cdot(n+k)+5\cdot(n+k)=(n+k)\cdot(m+5)\\\\5)\\\\a^{3} + a^{2} -5a-5=(a^{3} + a^{2} )-(5a+5)=\\\\=a^{2} \cdot(a+1)-5\cdot(a+1)=(a+1)\cdot(a^{2} -5)$

$\displaystyle\bf\\6)\\\\b^{3} +3b^{2} +2b+6=(b^{3} +2b)+(3b^{2} +6)=\\\\=b\cdot(b^{2} +2)+3\cdot(b^{2} +2)=(b^{2} +2)\cdot(b+3)\\\\7)\\\\b^{3} -3b^{2} -2b+6=(b^{3} -2b)-(3b^{2} -6)=\\\\=b\cdot(b^{2} -2)-3\cdot(b^{2} -2)=(b^{2} -2)\cdot(b-3)\\\\8)\\\\y^{3} -by^{2} +3b^{2} y-3b^{3} =(y^{3} -by^{2} )+(3b^{2} y-3b^{3} )=\\\\=y^{2} \cdot(y-b)+3b^{2} \cdot(y-b)=(y-b)\cdot(y^{2} +3b^{2} )\\\\9)$

$\displaystyle\bf\\35x^{2} +35xy+20x+20y=(35x^{2} +35xy)+(20x+20y)=\\\\35x\cdot(x+y)+20\cdot(x+y)=(x+y)\cdot(35x+20)=5\cdot(x+y)\cdot(7x+4)\\\\10)\\\\8x^{2} -12xy+14xz-21yz=(8x^{2} -12xy)+(14xz-21yz)=\\\\=4x\cdot(2x-3y)+7z\cdot(2x-3y)=(2x-3y)\cdot(4x+7z)$