`7,` `x^4-5x^2+4=x^4-x^2-4x^2+4=x^2(x^2-1)-4(x^2-1)=(x^2-4)(x^2-1)=(x-2)(x+2)(x-1)(x+1)`
`8,` `49-x^2+2xy-y^2=7^2-(x^2-2xy+y^2)=7^2-(x-y)^2=(7+x-y)(7-x+y)`
`9,` `x^3-4x^2+5x-2=x^3-x^2-3x^2+3x+2x-2=x^2(x-1)-3x(x-1)+2(x-1)=(x-1)(x^2-3x+2)=(x-1)(x^2-x-2x+2)=(x-1)[x(x-1)-2(x-1)]=(x-1)(x-1)(x-2)=(x-2)(x-1)^2`
`10,` `(x+1)(x+3)(x+5)(x+7)+15=(x+1)(x+7)(x+3)(x+5)+15=(x^2+7x+x+7)(x^2+5x+3x+15)+15=(x^2+8x+7)(x^2+8x+15)+15`
Đặt `a=x^2+8x+7` có:
`a.(a+8)+15=a^2+8a+15=a^2+3a+5a+15=a(a+3)+5(a+3)=(a+3)(a+5)`
Hay `(x^2+8x+7+3)(x^2+8x+7+5)=(x^2+8x+10)(x^2+8x+12)=(x^2+8x+10)(x^2+6x+2x+12)=(x^2+8x+10)[x(x+6)+2(x+6)]=(x^2+8x+10)(x+2)(x+6)`
`11,` `45+x^3-5x^2-9x=x^3-5x^2-9x+45=x^2(x-5)-9(x-5)=(x-5)(x^2-9)=(x-5)(x+3)(x-3)`
`12,` `2x^3+6xy-x^2z-3yz=2x^3-x^2z+6xy-3yz=x^2(2x-z)+3y(2x-z)=(2x-z)(x^2+3y)`