Question

Разложите на множители трехчлены, выделив предварительно квадрат двучлена:

Answer 1

$\displaystyle\bf\\1)\\\\x^{2} -2x-3=(x^{2} -2x+1)-1-3=(x^{2} -2x+1)-4=\\\\=(x-1)^{2} -2^{2} =(x-1-2)\cdot(x-1+2)=(x-3)\cdot(x+1)\\\\2)\\\\x^{2} +4x-5=(x^{2} +4x+4)-4-5=(x^{2} +4x+4)-9=\\\\=(x+2)^{2} -3^{2} =(x+2-3)\cdot(x+2+3)=(x-1)\cdot(x+5)\\\\3)\\\\x^{2} +6x-7=(x^{2} +6x+9)-9-7=(x^{2} +6x+9)-16=\\\\=(x+3)^{2} -4^{2} =(x+3-4)\cdot(x+3+4)=(x-1)\cdot(x+7)\\\\4)\\\\x^{2} -8x-9=(x^{2} -8x+16)-16-9=(x^{2} -8x+16)-25=\\\\=(x-4)^{2} -5^{2} =(x-4-5)\cdot(x-4+5)=(x-9)\cdot(x+1)$