5.4-bo'lim : ko'phadlarning nollarini topish
2. Quyidagi ko‘phadning barcha nollarini toping.
\[P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4\]Yechim
Bizga 4 va 1 ning barcha omillari kerak bo'ladi.
\[\begin{align*}4 & : \pm 1, \pm 2, \pm 4\\ 1 & : \pm 1\end{align*}\]2-qadam
Bu ko'phad uchun barcha mumkin bo'lgan ratsional nollarning ro'yxati.
\[\frac{{ \pm 1}}{{ \pm 1}} = \pm 1\hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} = \pm 2\hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} = \pm 4\]Demak, ko'phad uchun jami 6 ta nol mavjud.
3-qadam
Endi biz sintetik bo'linish ishini boshlashimiz kerak. Biz birinchi navbatda "kichik" butun sonlardan boshlaymiz.
\[\,\,\begin{array}{r|rrrrl} {} & 1 & -3 & -5 & 3 & 4 \\ \hline -1 & 1 & -4 & -1 & 4 & 0 =P\left( -1 \right)=0!! \\ \end{array}\]Endi biz \(x = - 1\) nol ekanligini bilamiz va ko'phadni quyidagicha yozishimiz mumkin:
\[P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4 = \left( {x + 1} \right)\left( {{x^3} - 4{x^2} - x + 4} \right)\]4-qadam
Shunday qilib, endi biz \(Q\left( x \right) = {x^3} - 4{x^2} - x + 4\) yordamida jarayonni davom ettirishimiz kerak. Ushbu ko'phadning mumkin bo'lgan nollari asl ko'phad bilan bir xil, shuning uchun biz ularni qayta yozmaymiz.
Mana bu \(Q\left( x \right)\) uchun sintetik bo'linish ishi. \[\begin{array}{r|rrrl} {} & 1 & -4 & -1 & 4 \\ \hline -1 & 1 & -5 & 4 & 0 =Q\left( -1 \right)=0!! \\ \end{array}\]Shunday qilib, \(x = - 1\) ham \(Q\left( x \right)\) ning nolga teng va \(Q\left( x \right)\) ning faktorlangan shakli:
\[Q\left( x \right) = {x^3} - 4{x^2} - x + 4 = \left( {x + 1} \right)\left( {{x^2} - 5x + 4} \right)\]Bu shuni anglatadiki, asl ko'phadning faktorlangan shakli endi,
\[P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4 = \left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - 5x + 4} \right) = {\left( {x + 1} \right)^2}\left( {{x^2} - 5x + 4} \right)\]5-qadam
Biz kvadratik ko'phadga tushdik va shuning uchun biz asl ko'phadning to'liq faktorlangan shaklini olish uchun buni faktorlar qilishimiz mumkin. Bu,
\[P\left( x \right) = {x^4} - 3{x^3} - 5{x^2} + 3x + 4 = {\left( {x + 1} \right)^2}\left( {x - 4} \right)\left( {x - 1} \right)\]6-qadam
To'liq faktorlangan shakldan biz asl ko'phad uchun quyidagi nollar to'plamini olamiz.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1\,\,\,({\mbox{multiplicity 2}})\hspace{0.25in}\hspace{0.25in}x = 1\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}x = 4}}\]