Tenglamani yeching: (x2+x+1)(x10+x9+ … +x+1) = (x6+x5+… +x+1)2
Yechilishi: Tenglamaning ikkala qismini (x-1)2 ga ko’paytirsak, (x3-1)(x11-1) = (x7-1)2 tenglamaga ega bo’lamiz. Qavslarni ochsak, x14-x11-x3+1 = x14-2x7+1 bundan esa x11-2x7+x3=0
x3(x8-2x4+1) =0 , x3(x4-1)2 =0; x1=0, x2=1, x3= -1.
Endi, tenglamaning ikkala qismini (x-1)2 ga ko’paytirganimizda hosil bo’lishi mumkin bo’lgan ildizni tekshirish kerak. Bu ildiz, x=1. O’rniga qo’yib tekshirib x=1 chet ildiz ekanini aniqlaymiz.
Javob: x1=0, x2 =-1.