Bu ko'phad uchun barcha mumkin bo'lgan ratsional nollarning ro'yxati.

\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2\\ \\ \frac{{ \pm 1}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 4}} & = \pm 1\\ \\ \frac{{ \pm 1}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 8}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 2}}\end{align*}\]

Demak, ko'phad uchun jami 12 ta nol mavjud.

Endi biz sintetik bo'linish ishini boshlashimiz kerak. Biz birinchi navbatda "kichik" butun sonlardan boshlaymiz.

\[\begin{array}{r|rrrrrl} {} & 8 & 36 & 46 & 7 & -12 & -4 \\ \hline -1 & 8 & 28 & 18 & -11 & -1 & -3 =g\left( -1 \right)\ne 0 \\ 1 & 8 & 44 & 90 & 97 & 85 & 81 =g\left( 1 \right)\ne 0 \\ \end{array}\]

E'tibor bering, yuqorida sanab o'tilgan ikkita funktsiyani baholash uchun qarama-qarshi belgilar mavjud. Eslatib o'tamiz, bu biz ular orasida nol borligini bilamiz.

Shunday qilib, keling, ro'yxatimizdagi ba'zi kasrlarni ko'rib chiqamiz va ularni sintetik bo'linish jadvalida sinab ko'raylik. Biz eng kichik maxrajli kasrlardan boshlaymiz.

\[\begin{array}{r|rrrrrl} {} & 8 & 36 & 46 & 7 & -12 & -4 \\ \hline -\frac{1}{2} & 8 & 32 & 30 & -8 & -8 & 0 =g\left( -1 \right)=0!! \\ \end{array}\]

Endi biz \(x = - \frac{1}{2}\) nol ekanligini bilamiz va ko'phadni quyidagicha yozishimiz mumkin:

\[g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 = \left( {x + \frac{1}{2}} \right)\left( {8{x^4} + 32{x^3} + 30{x^2} - 8x - 8} \right)\]

Shunday qilib, endi biz \(Q\left( x \right) = 8{x^4} + 32{x^3} + 30{x^2} - 8x - 8\) yordamida jarayonni davom ettirishimiz kerak. Mana \(Q\left( x \right)\) ning barcha mumkin bo'lgan nollari ro'yxati.

\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 1}} & = \pm 8\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 2}} & = \pm 4\\ \\ \frac{{ \pm 1}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 4}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 4}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 4}} & = \pm 2\\ \\ \frac{{ \pm 1}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 8}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 4}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 8}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 8}} & = \pm 1\end{align*}\]

Shunday qilib, bizda 14 ta mumkin bo'lgan nollar ro'yxati mavjud (ko'p takrorlashlar), lekin biz allaqachon \( \pm 1\) asl ko'phadning nollari bo'la olmasligini va shuning uchun \(Q\left( x \right)\) ning nollari bo'lishi mumkin emasligini allaqachon isbotlaganimizni unutmang.

Mana bu \(Q\left( x \right)\) uchun sintetik bo'linish ishi.

\[\begin{array}{r|rrrrl} {} & 8 & 32 & 30 & -8 & -8 \\ \hline -2 & 8 & 16 & -2 & -4 & 0 =Q\left( -2 \right)=0!! \\ \end{array}\]

Shunday qilib, \(x = - 2\) \(Q\left( x \right)\) ning nolga teng va \(Q\left( x \right)\) ning faktorlangan shakli:

\[Q\left( x \right) = 8{x^4} + 32{x^3} + 30{x^2} - 8x - 8 = \left( {x + 2} \right)\left( {8{x^3} + 16{x^2} - 2x - 4} \right)\]

Bu shuni anglatadiki, asl ko'phadning faktorlangan shakli endi,

\[g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 = \left( {x + \frac{1}{2}} \right)\left( {x + 2} \right)\left( {8{x^3} + 16{x^2} - 2x - 4} \right)\]

Shunday qilib, biz sintetik bo'linishni davom ettirishimiz kerak. Bu safar biz buni \(P\left( x \right) = 8{x^3} + 16{x^2} - 2x - 4\) ko'phadida qilamiz.

Buning mumkin bo'lgan nollari asl ko'phad bilan bir xil, shuning uchun biz ularni bu erda yozmaymiz. Shunga qaramay, biz \( \pm 1\) asl ko'phadning nolga teng bo‘lishi va \(P\left( x \right)\) ning ham nolga teng bo‘lishi mumkin emasligini allaqachon isbotladik.

Mana \(P\left( x \right)\) uchun sintetik bo'lim.

\[\begin{array}{r|rrrl} {} & 8 & 16 & -2 & -4 \\ \hline -2 & 8 & 0 & -2 & 0 =P\left( -2 \right)=0!! \\ \end{array}\]

Shunday qilib, \(x = - 2\) \(P\left( x \right)\) ning nolga teng va \(P\left( x \right)\) ning faktorlangan shakli:

\[P\left( x \right) = 8{x^3} + 16{x^2} - 2x - 4 = \left( {x + 2} \right)\left( {8{x^2} - 2} \right) = 8\left( {x + 2} \right)\left( {{x^2} - \frac{1}{4}} \right)\]

Keyingi bosqichni biroz osonlashtirish uchun biz kvadratdan 8 ni ajratdik.

Asl ko'phadning faktorlangan shakli endi,

\[g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 = 8\left( {x + \frac{1}{2}} \right){\left( {x + 2} \right)^2}\left( {{x^2} - \frac{1}{4}} \right)\]

Biz kvadratik ko'phadga tushdik va shuning uchun biz asl ko'phadning to'liq faktorlangan shaklini olish uchun buni faktorlar qilishimiz mumkin. Bu,

\[\begin{align*}g\left( x \right) = 8{x^5} + 36{x^4} + 46{x^3} + 7{x^2} - 12x - 4 & = 8\left( {x + \frac{1}{2}} \right){\left( {x + 2} \right)^2}\left( {x + \frac{1}{2}} \right)\left( {x - \frac{1}{2}} \right)\\ & = 8{\left( {x + \frac{1}{2}} \right)^2}{\left( {x + 2} \right)^2}\left( {x - \frac{1}{2}} \right)\end{align*}\]

To'liq faktorlangan shakldan biz asl ko'phad uchun quyidagi nollar to'plamini olamiz.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{1}{2}\,\,\,({\mbox{multiplicity 2}})\hspace{0.25in}\hspace{0.25in}x = - 2\,\,\,({\mbox{multiplicity 2}})\hspace{0.25in}\hspace{0.25in}x = \frac{1}{2}}}\]

E'tibor bering, bu muammo juda uzoq va tartibsiz edi. Bu muammoning maqsadi haqiqatan ham ular qanchalik uzoq va tartibsiz bo'lishi mumkinligini ko'rsatish edi. Axloqiy, agar mavjud bo'lsa, biz odatda orqaga o'tiramiz va haqiqatan ham bunday muammolarni muntazam ravishda hal qilishimiz shart emas deb umid qilamiz. Ular juda uzun va ular bilan xato qilish juda oson.