5.4-bo'lim : ko'phadlarning nollarini topish
3. Quyidagi ko‘phadning barcha nollarini toping.
\[A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24\]Yechim
Bizga -24 va 2 ning barcha omillari kerak bo'ladi.
\[\begin{align*} - 24 & : \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\\ 2 & : \pm 1, \pm 2\end{align*}\]2-qadam
Bu ko'phad uchun barcha mumkin bo'lgan ratsional nollarning ro'yxati.
\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 1}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 1}} & = \pm 6 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 1}} & = \pm 8\\ \\ & & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in}\,\,\,\,\frac{{ \pm 12}}{{ \pm 1}} & = \pm 12 & \hspace{0.25in}\,\,\,\,\,\frac{{ \pm 24}}{{ \pm 1}} & = \pm 24\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 2}} & = \frac{{ \pm 3}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 2}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 8}}{{ \pm 2}} & = \pm 4\\ \\ & & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in} & \hspace{0.25in}\,\,\,\,\frac{{ \pm 12}}{{ \pm 2}} & = \pm 6 & \hspace{0.25in}\,\,\,\,\,\frac{{ \pm 24}}{{ \pm 2}} & = \pm 12\end{align*}\]Demak, ko'phad uchun jami 20 ta nol mavjud.
3-qadam
Endi biz sintetik bo'linish ishini boshlashimiz kerak. Biz birinchi navbatda "kichik" butun sonlardan boshlaymiz.
\[\begin{array}{r|rrrrl} {} & 2 & -7 & -2 & 28 & -24 \\ \hline -1 & 2 & -9 & 7 & 21 & -45=A\left( -1 \right)\ne 0 \\ 1 & 2 & -5 & -7 & 21 & -3=A\left( 1 \right)\ne 0 \\ -2 & 2 & -11 & 20 & -12 & 0 =A\left( -2 \right)=0!! \\ \end{array}\]Endi biz \(x = - 2\) nol ekanligini bilamiz va ko'phadni quyidagicha yozishimiz mumkin:
\[A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24 = \left( {x + 2} \right)\left( {2{x^3} - 11{x^2} + 20x - 12} \right)\]4-qadam
Shunday qilib, endi biz \(Q\left( x \right) = 2{x^3} - 11{x^2} + 20x - 12\) yordamida jarayonni davom ettirishimiz kerak. Mana \(Q\left( x \right)\) ning barcha mumkin bo'lgan nollari ro'yxati.
\[\begin{align*}\frac{{ \pm 1}}{{ \pm 1}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 1}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 1}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 1}} & = \pm 4 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 1}} & = \pm 6 & \hspace{0.25in}\frac{{ \pm 12}}{{ \pm 1}} & = \pm 12\\ \\ \frac{{ \pm 1}}{{ \pm 2}} & = \frac{{ \pm 1}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 2}}{{ \pm 2}} & = \pm 1 & \hspace{0.25in}\frac{{ \pm 3}}{{ \pm 2}} & = \frac{{ \pm 3}}{{ \pm 2}} & \hspace{0.25in}\frac{{ \pm 4}}{{ \pm 2}} & = \pm 2 & \hspace{0.25in}\frac{{ \pm 6}}{{ \pm 2}} & = \pm 3 & \hspace{0.25in}\frac{{ \pm 12}}{{ \pm 2}} & = \pm 6\end{align*}\]Shunday qilib, bizda 16 ta mumkin bo'lgan nollar ro'yxati bor, lekin shuni yodda tutingki, biz \( \pm 1\) asl ko'phadning nolga teng bo'lishi va \(Q\left( x \right)\) ning nollari bo'lishi mumkin emasligini allaqachon isbotlaganmiz.
Bu \(Q\left( x \right)\) uchun sintetik bo'linish ishi.
\[\,\,\begin{array}{r|rrrl} {} & 2 & -11 & 20 & -12 \\ \hline -2 & 2 & -15 & 50 & -112 =Q\left( -2 \right)\ne 0 \\ 2 & 2 & -7 & 6 & 0 =Q\left( 2 \right)=0!! \\ \end{array}\]Shunday qilib, \(x = 2\) \(Q\left( x \right)\) ning nolga teng va \(Q\left( x \right)\) ning faktorlangan shakli:
\[Q\left( x \right) = 2{x^3} - 11{x^2} + 20x - 12 = \left( {x - 2} \right)\left( {2{x^2} - 7x + 6} \right)\]Bu shuni anglatadiki, asl ko'phadning faktorlangan shakli endi,
\[A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24 = \left( {x + 2} \right)\left( {x - 2} \right)\left( {2{x^2} - 7x + 6} \right)\]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,
\[A\left( x \right) = 2{x^4} - 7{x^3} - 2{x^2} + 28x - 24 = \left( {x + 2} \right)\left( {x - 2} \right)\left( {x - 2} \right)\left( {2x - 3} \right)=\left( {x + 2} \right){\left( {x - 2} \right)^2}\left( {2x - 3} \right)\]6-qadam
To'liq faktorlangan shakldan biz asl ko'phad uchun quyidagi nollar to'plamini olamiz.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 2\hspace{0.25in}\hspace{0.25in}x = \frac{3}{2}\hspace{0.25in}\hspace{0.25in}x = 2\,\,\,({\mbox{multiplicity 2}})}}\]