Materi kelas XI, materi kelas XI semester II, suku banyak, Uncategorized

Suku Banyak

A. Bentuk Umum

f(x)=\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}.

Selanjutnya,

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{f(x)=\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}}\\\hline \begin{aligned}a_{n}&\: \: \textrm{adalah koefisien dari} \: \: x^{n}\\ a_{n-1}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-1}\\ a_{n-2}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-2}\\ \vdots &\\ a_{2}&\: \: \textrm{adalah koefisien dari} \: \: x^{2}\\ a_{1}&\: \: \textrm{adalah koefisien dari} \: \: x^{1}\\ a_{0}&\: \: \textrm{adalah konstanta(suku tetap)} \\\end{aligned}&\begin{aligned}a_{n}\: &\: \neq 0\\ n:&\: \: \textrm{bilangan cacah},\\ :&\: \: \textrm{adalah derajat (pangkat)} \\ &\: \: \textrm{tertinggi dalam suku banyak} \\ &\: \: \textrm{tersebut}\\ &\\ &\end{aligned}\\\hline \multicolumn{2}{|l|}{\begin{aligned}&\\ \textrm{Nilai}\: &\: \textrm{suku banyak}\: \: f(x)\: \: \textrm{berderajat n saat}\: \: x = k\: \: \textrm{adalah}\: \: f(k).\\ &\textrm{Jika}\: \: f(k)=0\: \: \textrm{maka}\: \: x = k\: \: \textrm{akar dari}\: \: f(x),\\ &\textrm{dan}\: \: (x-k)\: \: \textrm{faktor dari}\: \: f(x)\\ &\end{aligned}}\\\hline \end{array}.

B. Pembagian dan Nilai Suku Banyak

Ada 2 cara, yaitu:

  1. Pembagian biasa (cara bersusun atau bentuk panjang)
  2. Metode pembagian sintetis atau pembagian cara Horner dan atau Horner-Kino

Perhatikanlah ilustrasi berikut

\begin{array}{|c|c|c|c|c|c|c|}\hline 9801&=&9&\times &1089&+&0\\\cline{1-1}\cline{3-3}\cline{5-5}\cline{7-7} &&&&&&\\ \Downarrow &&\Downarrow &&\Downarrow &&\Downarrow \\ &&&&&&\\ \textrm{yang dibagi}&&\textrm{pembagi}&&\textrm{hasil bagi}&&\textrm{sisa pembagian}\\ &&&&&&\\\hline \end{array}.

Berikut contoh pembagian cara biasa,

\begin{array}{rlllll}\\ \multicolumn{6}{l}{\qquad\qquad\quad \begin{aligned}\textrm{hasil}&\: \: \textrm{bagi}\\ &\Downarrow\\ & \end{aligned}}\\ &x^{2}+5x+3&&&&\\\cline{2-2} x-1\diagup &x^{3}+4x^{2}-2x+4& &\cdots &\Rightarrow&\textrm{yang dibagi}\\ &x^{3}-x^{2}&-&&&\\\cline{2-2} \Uparrow\qquad &\qquad 5x^{2}-2x&&&&\\ &\qquad 5x^{2}-5x&-&&&\\\cline{2-2} \textrm{pembagi}\quad &\qquad\qquad\: \: \: 3x+4&&&&\\ &\qquad\qquad\: \: \: 3x-3&-&&&\\\cline{2-2} &\qquad\qquad\qquad\quad 7& &\cdots &\Rightarrow&\textrm{sisa pembagian}\\ \end{array}.

Jika contoh pembagian cara biasa di atas kita tampilkan dengan model Horner, maka akan berupa ilustrasi sebagai berikut:

\begin{array}{l|l|lllll} \text{x = 1}&&\text{1}&\text{4}&\text{-2}&\text{4}&\\\cline{1-1} \multicolumn{2}{l|}{.}&&&&&\\ \multicolumn{2}{c|}{.}&&1&5&3&+\\\cline{3-6} \multicolumn{3}{r}{.}&&&&\\ \multicolumn{3}{r}{1}&5&3&\fbox{7}&\\ \end{array}.

Perlu diingat di sini bahwa jika f(x) adalah suku banyak berderajat n dan p(x) suku banyak berderajat k dengan  k < n , maka akan menghasilkan hasil bagi h(x) dengan sisa s(x),

f(x) = p(x) . h(x) + s(x)

dengan derajat h(x) adalah (n – k) dan derajat s(x) kurang dari k.

C. Kesamaan Suku Banyak

Jika dua suku banyak dengan sebuah variabel identik, koefisien-koefisien dari variabel yang bersesuaian derajatnya akan sama.

D. Teorema Sisa dan Teorema Faktor

\begin{aligned}\textrm{Perhatikan skema berikut}:&\\ &f(x)=(x-k).h(x)+s(x)\begin{cases} f(x) &: \text{ suku banyak} \\ (x-k) &: \text{ pembagi atau }p(x) \\ h(x) &: \text{ hasil bagi }\\ s(x) &: \text{ sisa pembagian } \end{cases} \end{aligned}.

Selanjutnya,

\begin{array}{|c|l|}\hline \multicolumn{2}{|c|}{f(x)=p(x)\times h(x)+s(x)}\\\hline \textrm{Teorema Sisa}&\textrm{Teorema Faktor}\\\hline \begin{cases} f(x) &:(x-k) \text{ maka} \: \: s(x)=f(k)=c=\textrm{konstanta} \\ f(x) &:(x+k) \text{ maka } \: \: s(x)=f(-k)=c=\textrm{konstanta} \\ f(x) &:(ax-b)\text{ maka }\: \: s(x)=f\left ( \displaystyle \frac{b}{a} \right )\\ f(x) &:(ax+b)\text{ maka }\: \: s(x)=f\left ( -\displaystyle \frac{b}{a} \right )\\ f(x) &:(ax^{2}+bx+c) \text{ maka }\: \: s(x)=px+q\\ f(x) &:(ax^{3}+bx^{2}+cx+d)\text{ maka }\: \: s(x)=px^{2}+qx+r\\ &dst \end{cases}&\begin{aligned}&\textrm{Jika}\: \: f(k)=0,\\ &\textrm{maka}\: \: (x-k)\\ & \textrm{faktor dari}\: \: f(x)\\ &\textrm{dan persamaan}\\ &\textrm{menjadi:}\\ &f(x)=(x-k).h(x)\\ &\\ & \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Sebutkanlah nama variabel (peubah), derajat, dan koefisien-koefisien dari tiap-tiap suku banyak berikut}\\ &\textrm{a})\quad 2x^{3}-5x^{2}-10x+6\\ &\textrm{b})\quad x^{3}+4x-2\\ &\textrm{c})\quad 3-2m+7m^{2}-m^{3}\\ &\textrm{d})\quad z^{12}+4z^{8}-z^{6}+z^{4}-5z^{2}+z+6 \end{array}\\\\\\ \textrm{Jawab}:\\\.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a})\quad &\textrm{adalah suku banyak dalam variabel}\: \: x\\ &\textrm{berderajat 3, dengan}\\ &\textrm{koefisien}\: \: x^{3}\: \: \textrm{adalah}\: \: 2\\ &\textrm{koefisien}\: \: x^{2}\: \: \textrm{adalah}\: \: -5\\ &\textrm{koefisien}\: \: x\: \: \textrm{adalah}\: \: -10\: \: \textrm{dan}\\ &\textrm{suku tetapnya(konstanta) adalah}\: \: 6 \end{aligned}&\begin{aligned}\textrm{b})\quad &\textrm{adalah suku banyak dalam variabel}\: \: x\\ &\textrm{berderajat 3, dengan}\\ &\textrm{koefisien}\: \: x^{3}\: \: \textrm{adalah}\: \: 1\\ &\textrm{koefisien}\: \: x^{2}\: \: \textrm{adalah}\: \: 0\\ &\textrm{koefisien}\: \: x\: \: \textrm{adalah}\: \: 4\: \: \textrm{dan}\\ &\textrm{suku tetapnya(konstanta) adalah}\: \: -2\end{aligned}\\\hline \begin{aligned}\textrm{c})\quad &\textrm{adalah suku banyak dalam variabel}\: \: m\\ &\textrm{berderajat 3, dengan}\\ &\textrm{koefisien}\: \: m^{3}\: \: \textrm{adalah}\: \: -1\\ &\textrm{koefisien}\: \: m^{2}\: \: \textrm{adalah}\: \: 7\\ &\textrm{koefisien}\: \: m\: \: \textrm{adalah}\: \: -2\: \: \textrm{dan}\\ &\textrm{suku tetapnya(konstanta) adalah}\: \: 3\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{d})\quad &\textrm{adalah suku banyak dalam variabel}\: \: z\\ &\textrm{berderajat 3, dengan}\\ &\textrm{koefisien}\: \: z^{12}\: \: \textrm{adalah}\: \: 1\\ &\textrm{koefisien}\: \: z^{11}\: \: \textrm{adalah}\: \: 0\\ &\textrm{koefisien}\: \: z^{10}\: \: \textrm{adalah}\: \: 0\: \: \textrm{dan}\\ &\vdots \\ &\textrm{koefisien}\: \: z^{2}\: \: \textrm{adalah}\: \: -5\\ &\textrm{koefisien}\: \: z\: \: \textrm{adalah}\: \: 1\\ &\textrm{suku tetapnya(konstanta) adalah}\: \: 6 \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah banyaknya variabel dan derajat dari tiap-tiap suku banyak berikut}\\ &\textrm{a})\quad x^{5}y+xy^{3}+4x-5y+12\\ &\textrm{b})\quad (2x-y)^{4}-(2y-3z)^{3}+(2z-x)^{6}\\ &\textrm{c})\quad a^{5}b^{5}-a^{4}b^{3}+a^{3}b^{2}+a+b^{2}-6\\ &\textrm{d})\quad p^{4}+q^{4}+r^{4}+3pq-5pr+6qr+4 \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\textrm{a})\quad &\textrm{adalah suku banyak dalam 2 variabel, yaitu}\: \: x\: \: \textrm{dan}\: \: y\\ &\textrm{berderajat 5 dalam variabel}\: \: x\\ &\textrm{dan berderajat 3 dalam variabel}\: \: y\\ & \end{aligned}&\begin{aligned}\textrm{b})\quad &\textrm{adalah suku banyak dalam 3 variabel, yaitu}\: \: x,\: y\: \: \textrm{dan}\: \: z\\ &\textrm{berderajat 6 dalam variabel}\: \: x\\ &\textrm{berderajat 4 dalam variabel}\: \: y\\ &\textrm{dan berderajat 6 dalam variabel}\: \: z \end{aligned}\\\hline \begin{aligned}\textrm{c})\quad &\textrm{adalah suku banyak dalam 2 variabel, yaitu}\: \: a\: \: \textrm{dan}\: \: b\\ &\textrm{berderajat 5 dalam variabel}\: \: a\\ &\textrm{dan berderajat 5 dalam variabel}\: \: b\\ & \end{aligned}&\begin{aligned}\textrm{d})\quad &\textrm{adalah suku banyak dalam 3 variabel, yaitu}\: \: p,\: q\: \: \textrm{dan}\: \: r\\ &\textrm{berderajat 4 dalam variabel}\: \: p\\ &\textrm{berderajat 4 dalam variabel}\: \: q\\ &\textrm{dan berderajat 4 dalam variabel}\: \: r \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Sederhanakanlah bentuk berikut}!\\ &\textrm{a})\quad (x^{3}+5x^{2}-2x+1)+(x^{3}+x+4)\\ &\textrm{b})\quad (2x^{4}-3x^{2}+4x+3)+(x^{4}-2x^{3}-x-8)\\ &\textrm{c})\quad (x^{5}-7x^{3}+3x)-(3-2x+5x^{2}-2x^{3})\\ &\textrm{d})\quad (3x^{4}+5x^{2}-x)-(x^{4}-2x^{3}+x^{2}-3)\\ &\textrm{e})\quad (2-3x+4x^{2}-5x^{4})-(x^{4}-3x^{3}+5x^{2}+x+2)\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\textrm{a})\quad (x^{3}+5x^{2}-2x+1)+(x^{3}+x+4)=2x^{3}+5x^{2}-x+5\\ &\textrm{b})\quad (2x^{4}-3x^{2}+4x+3)+(x^{4}-2x^{3}-x-8)=3x^{3}-2x^{3}-3x^{2}+3x-5\\ &\textrm{c})\quad (x^{5}-7x^{3}+3x)-(3-2x+5x^{2}-2x^{3})=...x^{5}-...x^{3}-...x^{2}+...x-...\\ &\textrm{d})\quad (3x^{4}+5x^{2}-x)-(x^{4}-2x^{3}+x^{2}-3)=...x^{4}+...x^{3}+...x^{2}-x+...\\ &\textrm{e})\quad (2-3x+4x^{2}-5x^{4})-(x^{4}-3x^{3}+5x^{2}+x+2)=-...x^{4}+...x^{3}-...x^{2}-...x \end{aligned}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah koefisien dari}\\ &\textrm{a})\quad x^{3}\: \: \textrm{pada}\: \: (2x^{2}-1)(x^{2}-2x+1)\\ &\textrm{b})\quad x^{2}\: \: \textrm{pada}\: \: (x^{2}-1)(1-x)(2x+3)\\ &\textrm{c})\quad x^{5}\: \: \textrm{pada}\: \: (x^{2}-2x)^{2}(2x+1)^{3}\\ &\textrm{d})\quad m^{3}\: \: \textrm{pada}\: \: (m-1)(m-2)(2m+3)m^{2}\\ &\textrm{e})\quad y^{3}\: \: \textrm{pada}\: \: (y-1)^{2}(y+2)(3x+1)\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textrm{yang dibahas hanya no 4.a) saja}\\\\ \begin{aligned}\textrm{a})\quad (2x^{2}-1)(x^{2}-2x+1)&=2x^{4}-4x^{3}+2x^{2}-x^{2}+2x-1\\ &=2x^{4}-4x^{3}+x^{2}+2x-1\\ &\: \: \textrm{sehingga koefisien dari}\: \: x^{3}\: \: \textrm{adalah -4} \end{aligned}.

\begin{array}{ll}\\ 5.&\textrm{Dengan metode Horner, tentukanlah nilai suku banyak berikut ini}!\\ &\textrm{a})\quad 4x^{4}-7x^{3}+8x^{2}-2x+3\: \: \: \textrm{jika}\: \: x=2\\ &\textrm{b})\quad 2x^{5}+3x^{3}-x+1\: \: \: \textrm{jika}\: \: x=-3\\ &\textrm{c})\quad 2x^{3}+x^{2}-2x+3\: \: \: \textrm{jika}\: \: x=\displaystyle \frac{1}{3}\\ &\textrm{d})\quad 5x^{4}+2x^{2}-x+1\: \: \: \textrm{jika}\: \: x=0,5\\ &\textrm{e})\quad (3x-1)^{2}(x^{2}-2)\: \: \: \textrm{jika}\: \: x=\displaystyle \frac{3}{2}\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textrm{yang dibahas hanya no. 5.a) saja}\\\\ \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{f(x)=4x^{4}-7x^{3}+8x^{2}-2x+3}\\\hline \textrm{Dengan Horner}&\textrm{Substitusi (hanya sebagai pembanding)}\\\hline \begin{aligned}&\\ &\begin{array}{l|l|lrrrrl} \multicolumn{2}{l}{.}&a_{4}&a_{3}&a_{2}&a_{1}&a_{0}\\ \multicolumn{2}{l}{.}&&&&&\\ \text{x = 2}&&\text{4}&\text{-7}&\text{8}&\text{-2}&3&\\\cline{1-1} \multicolumn{2}{l|}{.}&&&&&\\ \multicolumn{2}{c|}{.}&&8&2&20&36&+\\\cline{3-7} \multicolumn{3}{r}{.}&&&&\\ \multicolumn{3}{r}{4}&1&10&18&\fbox{39}\\ \end{array}\\ & \end{aligned}&\begin{aligned}f(2)&=4(2)^{4}-7(2)^{3}+8(2)^{2}-2(2)+3\\ &=64-56+32-4+3\\ &=39\\ \textrm{Sebagai}&\: \: \textrm{catatan bahwa}:\\ &\: \textrm{Polinom}\: \: f(x)\: \: \textrm{tersebut di atas jika dibagi}\: \: (x-2)\\ &\: \textrm{akan bersisa 39} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 6.&\textrm{Hitunglah nilai}\: \: a,\: b,\: c,\: \: \textrm{dan}\: \: d,\: \: \textrm{jika}\\ &\textrm{a})\quad -3x+4\equiv a(x-7)-b(2x-3)\\ &\textrm{b})\quad a(x-1)^{2}-b(x+4)\equiv 2x^{2}-5x-7\\ &\textrm{c})\quad 3x^{2}+2x-5\equiv (ax+1)(x+b)-c(x+1)+2(ab-c)\\ &\textrm{d})\quad x^{4}-8x^{3}+15x-20\equiv x^{4}+ax^{3}+(a+b)x^{2}+(2b-c)x+d\\ &\textrm{e})\quad \displaystyle \frac{a}{x-1}+\frac{b}{x+3}\equiv \displaystyle \frac{8}{x^{2}+2x-3}\\ &\textrm{f})\quad \displaystyle \frac{a}{x-1}+\frac{b}{x-4}\equiv \displaystyle \frac{3}{x-1}+\frac{20}{x-4}+\frac{x+17}{x^{2}-5x+4}\\ &\textrm{g})\quad \displaystyle \frac{5x-4}{x^{2}-1}\equiv \displaystyle \frac{a}{x-1}+\frac{b}{x+1}-\frac{3}{x^{2}-1}\\ &\textrm{h})\quad \displaystyle \frac{2x^{2}+x+2}{x^{3}-1}\equiv \displaystyle \frac{a}{x-1}+\frac{bx+c}{x^{2}+x+1}\\ &\textrm{i})\quad \displaystyle \frac{3x^{2}+2x-5}{x^{2}+5x+6}\equiv \displaystyle \frac{a(x-3)}{x+3}+\frac{b(x-5)}{x+2}+\frac{4c}{(x+2)(x+3)}\\ &\textrm{j})\quad x^{3}+ax^{2}+bx+c=0\: \: \textrm{dengan akar-akar}\: \: x_{1}=x_{2}=-1\: \: \textrm{dan}\: \: x_{3}=-3\\ &\textrm{k})\quad x^{3}+ax^{2}+bx+c=0\: \: \textrm{dengan akar-akar}\: \: 1,\: 2,\: \: \textrm{dan}\: \: 3 \end{array}.

Jawab:

Yang dibahas poin 6. d), yaitu:

\begin{aligned}x^{4}-8x^{3}+15x-20&\equiv x^{4}+ax^{3}+(a+b)x^{2}+(2b-c)x+d\\ \textrm{koefisien}\: \: x^{4}&:\: \: 1=1\\ \textrm{koefisien}\: \: x^{3}&:\: \: -8=a,\: \: \textrm{maka}\: \: a=-8\\ \textrm{koefisien}\: \: x^{2}&:\: \: 0=a+b,\: \: \textrm{maka}\: \: b=-a=-(-8)=8\\ \textrm{koefisien}\: \: x^{1}&:\: \: 15=2b-c,\: \: \textrm{maka}\: \: c=2b-15=2(8)-15=1\\ \textrm{koefisien}\: \: x^{0}&:\: \: -20=d,\: \: \textrm{maka}\: \: d=-20\\ \end{aligned}.

\begin{array}{ll}\\ 7.&\textrm{Tentukanlah hasil bagi dan sisanya}!\\ &\begin{array}{ll}\\ \textrm{a})\quad (3x^{3}-2x^{2}+x-4):(x-1)&\textrm{k})\quad (x^{7}+3x^{5}+1):(x^{2}-1)\\ \textrm{b})\quad (2x^{4}-3x^{3}+x^{2}-5x+3):(x-2)&\textrm{l})\quad (x^{4}-3x^{3}-5x^{2}+x-6):(x^{2}-x-2)\\ \textrm{c})\quad (3-x+4x^{2}-x^{3}):(x-3)&\textrm{m})\quad (2x^{4}-3x^{2}-x+2):(x^{2}-2x+1)\\ \textrm{d})\quad (x^{4}-x^{2}+11):(x+4)&\textrm{n})\quad (3x^{6}+4x^{4}-2x-1):(x-1)(x^{2}-4)\\ \textrm{e})\quad (x^{3}-10x+9):(x+5)&\textrm{o})\quad (x^{4}-4x^{3}+2x^{2}-x+1):(2x+1)(x^{2}-3x+2)\\ \textrm{f})\quad (2x^{3}-5x^{2}-11x+8):(3x+1)&\textrm{p})\quad (x^{7}-7x^{4}+3x):(x^{3}-4x)\\ \textrm{g})\quad (5x^{3}+11x^{2}+7x-4):(5x+1)&\textrm{q})\quad (2x^{3}+x^{2}-4x+5):(x^{2}+x+1)\\ \textrm{h})\quad (2x^{3}+5x^{2}-4x+5):(2x+3)&\textrm{r})\quad (2x^{4}+x^{3}-3x+6):(x^{2}+x+2)\\ \textrm{i})\quad (2x^{3}+7x^{2}-5x+4):(2x-1)&\textrm{s})\quad (x^{4}-3x^{2}+7x-4):(x^{2}-2x-1)\\ \textrm{j})\quad (6x^{3}-x^{2}+3):(2x-3)&\textrm{t})\quad (3x^{3}+4x-8):(3x^{2}+x+2) \end{array} \end{array}.

Jawab:

Yang dibahas No. 7. i)

\begin{array}{l|l|lrrrrl} \multicolumn{2}{l}{.}&a_{3}&a_{2}&a_{1}&a_{0}&\\ \multicolumn{2}{l}{.}&&&&&\\ x=\displaystyle \frac{1}{2}&&\text{2}&\text{7}&\text{-5}&\text{4}&&\\\cline{1-1} \multicolumn{2}{l|}{.}&&&&&\\ \multicolumn{2}{c|}{.}&&1&4&-\displaystyle \frac{1}{2}&&+\\\cline{3-7} \multicolumn{3}{r}{.}&&&&\\ \multicolumn{3}{r}{2}&8&-1&\boxed{\frac{7}{2}}&\\ \end{array} \quad \Rightarrow \begin{array}{|l|l|}\hline \textrm{Pembagi}&\begin{aligned}&\textrm{Sisa}\\ &s(x)=\displaystyle \frac{7}{2} \end{aligned}\\\cline{2-2} \begin{aligned}&2x-1=2(x-\frac{1}{2}) \end{aligned}&\begin{aligned}&\textrm{Hasil bagi}\\ &\displaystyle \frac{h(x)}{2}=\frac{2x^{2}+8x-1}{2}=x^{2}+4x-\frac{1}{2}\ \end{aligned}\\\hline \end{array}.

Sementara untuk No. 7. m)

\begin{array}{rr|rrrrrr} p=1&&2&0&-3&-1&2\\ &&&2&2&-1&-2&+\\\cline{3-7} q=1&&2&2&-1&-2&\fbox{0}\\ &&&2&4&3&\\\cline{3-7} \multicolumn{3}{r}{2}&4&3&\fbox{1} \end{array}\quad \Rightarrow \begin{array}{|l|l|}\hline \textrm{Pembagi}&\begin{aligned}&\textrm{Sisa}\\ &s_{2}(x-p)+s_{1}\\ &1(x-1)+0=x-1 \end{aligned}\\\cline{2-2} \begin{aligned}(x-p)(x-q)&=(x-1)(x-1)\\ &=(x-1)^{2} \end{aligned}&\begin{aligned}&\textrm{Hasil bagi}\\ &2x^{2}+4x+3 \end{aligned}\\\hline \end{array}.

Untuk jawaban No. 7 .m), jika ingin diselesaikan dengan cara pembagian Horner-Kino lihat di sini

E. Penyelesaian Persamaan Suku Banyak

  1. Akar-akar rasional suku banyak
  2. Jumlah dan hasil kali akar-akar suku banyak

Keterangan

\begin{array}{|c|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Akar-akar rasional}}&\textrm{Jumlah dan hasil kali}\\\hline \textrm{Faktor linear}&\textrm{Pemilihan}\: x=h&\textrm{Polinom berderajat}\: \: 3\\\hline \begin{aligned}&\textrm{Jika}\: \: (x-h)\: \textrm{adalah faktor}\\ &\textrm{dari}\: f(x)\: \textrm{maka}\: f(h)=0\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}1.\: \: &\textrm{Misalkan suku banyak} \\ &f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}.\\ &\textrm{Jika}\: \: r\: \: \textrm{adalah faktor dari}\: \: a_{0}\\ &\textrm{dan}\: \: s\: \: \textrm{faktor dari}\: \: a_{n},\: \textrm{maka akar rasional}\\ &\textrm{yang mungkin adalah}\: \: h=\displaystyle \frac{r}{s}.\\ 2.\: \: &\textrm{Jika ditemukan suatu akar}\: x=h_{1},\\ &\textrm{maka hasil bagi}\: \: f(x)\: \textrm{oleh}\: \: (x-h_{1})\\ &\textrm{adalah}\: \: h_{1}(x)\: \textrm{dan jika dimungkinkan}\\ &\textrm{carilah}\: \: x=h_{2}\: \textrm{demikian seterusnya} \end{aligned}&\begin{aligned}&\textrm{Jika}\: \: x_{1},\: x_{2},\: \textrm{dan}\: \: x_{3}\\ &\textrm{adalah akar-akar persamaan}\\ &ax^{3}+bx^{2}+cx+d=0\\ &\textrm{maka},\\ &ax^{3}+bx^{2}+cx+d\equiv (x-x_{1})(x-x_{2})(x-x_{3})\\ &x^{3}+\displaystyle \frac{b}{a}x^{2}+\frac{c}{a}x+\frac{d}{a}\\ &\: \: \equiv x^{3}-(x_{1}+x_{2}+x_{3})x^{2}+(x_{1}x_{2}\\ &\: \: \: +x_{1}x_{3}+x_{2}x_{3})x-(x_{1}x_{2}x_{3})\\ & \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{LANJUTAN CONTOH SOAL}}.

\begin{array}{lll}\\ 8&\textrm{a})&\textrm{Diketahui akar-akar persamaan}\: \: x^{3}-12x^{2}+28x+p=0\: \: \textrm{adalah}\: \: x_{1},x_{2},\: \: \textrm{dan}\: \: x_{3}.\: \: \textrm{Jika}\: \: x_{1}=x_{2}+x_{3},\\ &&\textrm{tentukan nilai}\: \: p\: \: \textrm{dan akar-akarnya}!\\ &\textrm{b})&\textrm{Diketahui akar-akar persamaan}\: \: x^{3}-6x^{2}+ax-6=0\: \: \textrm{adalah}\: \: x_{1},x_{2},\: \: \textrm{dan}\: \: x_{3}.\: \: \textrm{Jika}\: \: x_{1},x_{2},x_{3}\\ &&\textrm{membentuk deret aritmetika, Carilah nilai}\: \: a\: \: \textrm{dan akar-akarnya}!\\ &\textrm{c})&\textrm{Diketahui akar-akar persamaan}\: \: x^{3}+qx^{2}-6x+8=0\: \: \textrm{adalah}\: \: x_{1},x_{2},\: \: \textrm{dan}\: \: x_{3}.\: \: \textrm{Jika}\: \: x_{1},x_{2},x_{3}\\ &&\textrm{membentuk deret geometri, Carilah nilai}\: \: q\: \: \textrm{dan akar-akarnya}!\\ &\textrm{d})&\textrm{Akar-akar persamaan}\: \: x^{3}-14x^{2}+px+q=0\: \: \textrm{merupakan deret geometri dengan rasio 2},\\ &&\textrm{tentukan harga}\: \: p+q!\\ &\textrm{e})&\textrm{Akar-akar persamaan}\: \: x^{4}-8x^{3}+ax^{2}+bx+c=0\: \: \textrm{merupakan deret aritmetika dengan beda 2},\\ &&\textrm{tentukan nilai}\: \: a^{2}-2ab+c^{2}!\\ &\textrm{f})&\textrm{Jika diketahui}\: \: x^{3}+px^{2}+qx+r=0\: \: \textrm{sama dengan akar-akar persamaan}\\ &&x^{3}+rx^{2}+(p-q+16)x+p+3q-r-15=0\: \: \textrm{dikurangai 1},\: \: \textrm{maka carilah nilai}\: \: p,\: q,\: r,\: \: \textrm{dan akar-akarnya} \end{array}.

Berikut pembahasan untuk No. 8 a).

\begin{array}{|l|l|}\hline \begin{aligned}x^{3}-12x^{2}+28x+p&=0\\ ax^{3}+bx^{2}+cx+d&=0\\ \bullet \quad x_{1}+x_{2}+x_{3}&=-\displaystyle \frac{b}{a}\\ x_{1}+x_{1}&=-\displaystyle \frac{-12}{1}\\ 2x_{1}&=12\\ x_{1}&=6 \end{aligned}&\begin{aligned}&\\ &\begin{array}{rr|rrrrrr}\\ x=6&&1&-12&28&p&\\ &&&&&&\\ &&&6&-36&-48&+\\\cline{3-6} \multicolumn{3}{r}{1}&-6&-8&\fbox{p - 48}&\Rightarrow &p-48=0\\ \multicolumn{3}{c}{.}&&&&&p=48 \end{array}\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\blacklozenge \quad \textrm{Cara lain}:&\\ &\begin{aligned}f(x)&=x^{3}-12x^{2}+28x+p&\\ f(6)&=6^{3}-12(6)^{2}+28(6)+p&=0\: \: \\ &=216-432+168+p&=0\: \: \\ &\qquad\qquad\qquad-48+p&=0\: \: \\ &\qquad\qquad\qquad\qquad\qquad p&=48 \end{aligned}\\ & \end{aligned}}\\\hline \end{array}.

Contoh soal:

\begin{array}{ll}\\ \fbox{1}.&\textrm{Jika}\: \: g(x)=2x^{3}+x^{2}-x+1,\: \: \textrm{maka}\: \: g(1)=\: \: \\ &\begin{array}{lll}\\ \textrm{a}.\quad -2&&\textrm{d}.\quad 2\\ \textrm{b}.\quad -1&\textrm{c}.\quad 1&\textrm{e}.\quad 3 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}g(x)&=2x^{3}+x^{2}-x+1\\ g(1)&=2(1)^{3}+(1)^{2}-(1)+1\\ &=2+1-1+1\\ &=3 \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Jika}\: \: p(y)=5y^{4}+2r^{2}y^{3}+y^{2}+1\: \: \textrm{dan}\: \: q(y)=4y^{5}+3ry^{2}-3y-1\: \: \\ &\textrm{serta}\: \: p(-1)=q(-1),\: \: \textrm{maka nilai}\: \: r\: \: \textrm{sama dengan}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{2}\: \: \textrm{dan}\: \: 3&&\textrm{d}.\quad -\displaystyle \frac{3}{2}\\ \textrm{b}.\quad \displaystyle -\frac{3}{2}\: \: \textrm{dan}\: \: 3&\textrm{c}.\quad \displaystyle \frac{3}{2}\: \: \textrm{dan}\: \: -3&\textrm{e}.\quad 3 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{c}\\\\ \begin{aligned}p(-1)&=q(-1)\\ 5(-1)^{4}+2r^{2}(-1)^{3}+(-1)^{2}+1&=4(-1)^{5}+3r(-1)^{2}-3(-1)-1\\ 5-2r^{2}+1+1&=-4+3r+3-1\\ 9-3r-2r^{2}&=0\\ \displaystyle \frac{(-6-2r)(-3+2r)}{2}&=0,\qquad \textrm{ingat pemfaktoran}\\ (-3-r)(-3+2r)&=0\\ r=-3\quad \vee \quad r&=\displaystyle \frac{3}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui}\: \: f(x)\: \: \textrm{berderajat}\: \: n. \: \: \textrm{Jika pembaginya berbentuk}\: \: \left ( ax^{2}+bx+c \right ),\\ &\textrm{dengan}\: \: a\neq 0,\: \: \textrm{maka hasil baginya berderajat}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad n-1&&\textrm{d}.\quad 3\\ \textrm{b}.\quad n-2&\textrm{c}.\quad n-3&\textrm{e}.\quad 2 \end{array}\end{array}\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}\textrm{Suku banyak (polinom)}&=\textrm{pembagi}\times \textrm{hasil bagi}+\textrm{sisa}\\ x^{n}+...&=\left ( ax^{2}+bx+c \right )\times \left ( x^{n-2}+... \right )+\left (mx+n \right ) \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Hasil bagi dan sisanya jika}\: \: \left (6x^{4}-3x^{2}+x-1 \right )\: \: \textrm{dibagi oleh}\: \: \left ( 2x-1 \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+\frac{1}{8}&\textrm{dan}&-\displaystyle \frac{7}{8}\\ \textrm{b}.\quad 3x^{3}+3x^{2}-\displaystyle \frac{3}{4}x+1&\textrm{dan}&-7\\ \textrm{c}.\quad x^{3}+\displaystyle \frac{3}{2}x^{2}-3x+\frac{1}{8}&\textrm{dan}&\displaystyle \frac{7}{8}\\ \textrm{d}.\quad x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+1&\textrm{dan}&\displaystyle \frac{1}{8}\\ \textrm{e}.\quad 3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x-\frac{1}{8}&\textrm{dan}&-\displaystyle \frac{7}{8}\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{array}{rr|rrrrrrr} x=\frac{1}{2}&&6&0&-3&1&-1\\ &&&3&\frac{3}{2}&-\frac{3}{4}&\frac{1}{8}&+&\\\cline{3-7} \multicolumn{3}{r}{6}&3&-\frac{3}{2}&\frac{1}{4}&\boxed{-\frac{7}{8}} \end{array}\begin{cases} \textrm{Hasil bagi}: & \displaystyle \frac{6x^{3}+3x^{2}-\frac{3}{2}x+\frac{1}{4}}{2}=3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+\displaystyle \frac{1}{8} \\ & \\ \textrm{Sisa bagi}: & -\displaystyle \frac{7}{8} \end{cases}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Hasil bagi dan sisanya jika}\: \: \left (x^{4}-x^{3}-x^{2}+x-1 \right )\: \: \textrm{dibagi oleh}\: \: \left ( x-2 \right )\left ( x+1 \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{2}+1&\textrm{dan}&2x+1\\ \textrm{b}.\quad x^{2}+1&\textrm{dan}&2x-1\\ \textrm{c}.\quad x^{2}-1&\textrm{dan}&2x+1\\ \textrm{d}.\quad x^{2}-1&\textrm{dan}&2x-1\\ \textrm{e}.\quad 2x^{2}-1&\textrm{dan}&x+1\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \textrm{Dengan menggunakan pembagian}\: \: \textbf{Horner-Kino}\: \: \textrm{akan diperoleh},\\\\ \begin{array}{rr|rrr|rrr} &&1&-1&-1&1&-1\\ 2&&\ast &\ast &2&0&2&\\\cline{3-7} 1&&\ast &1&0&1&\ast &+\\\cline{3-7} \multicolumn{3}{r}{1}&0&1&2&1\\\cline{6-7} \end{array}\quad \Rightarrow \begin{cases} \textrm{Suku banyak}: & f(x)=x^{4}-x^{3}-x^{2}+x-1 \\ \textrm{Pembagai}: & p(x)=(x-2)(x+1)=x^{2}-x-2 \\ &: 2\: \: \textrm{dari}\: -\frac{-2}{1},\: \: \textrm{sedang}\: \: 1=-\left ( \frac{-1}{1} \right )\\ \textrm{Hasil bagi}:&h(x)=x^{2}+1\\ \textrm{Sisa bagi}:&s(x)=2x+1 \end{cases} \\\\ \textrm{Sehingga},\\\\ x^{4}-x^{3}-x^{2}+x-1=\left ( x^{2}-x-2 \right )\left ( x^{2}+1 \right )+2x+1.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Diketahui bahwa}\: \: \displaystyle \frac{f(x)}{x-2}=h(x)+\displaystyle \frac{3}{x-2}\: \: \textrm{dan}\: \: \displaystyle \frac{f(x)}{x-1}=h(x)+\displaystyle \frac{2}{x-1}\: ,\\ &\textrm{jika}\: \: \displaystyle \frac{f(x)}{(x-2)(x-1)}=h(x)+\displaystyle \frac{s(x)}{(x-2)(x-1)}\: ,\: \: \textrm{maka}\: \: s(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x+1&&\textrm{d}.\quad 2x-1\\ \textrm{b}.\quad x+2&\textrm{c}.\quad 2x+1&\textrm{e}.\quad x-2\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}\displaystyle \frac{f(x)}{x-2}=h(x)+\displaystyle \frac{3}{x-2}\Rightarrow f(x)&=(x-2).h(x)+3\Rightarrow f(2)=3\\ \displaystyle \frac{f(x)}{x-1}=h(x)+\displaystyle \frac{2}{x-1}\Rightarrow f(x)&=(x-1).h(x)+2\Rightarrow f(1)=2\\ \displaystyle \frac{f(x)}{(x-2)(x-1)}&=h(x)+\displaystyle \frac{s(x)}{(x-2)(x-1)}\\ \textrm{maka}\qquad\qquad f(x)&=(x-2)(x-1).h(x)+s(x)\\ f(x)&=(x-2)(x-1).h(x)+px+q\\ f(2)&=2p+q=3\\ f(1)&=p+q=2,\\ \textrm{sehingga dengan }&\textrm{eliminasi akan diperoleh}\\ p&=1\quad \textrm{dan}\\ q&=1\\ \textrm{Jadi},\quad\qquad px+q&=x+1 \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: x^{4}+2mx-n\: \: \textrm{dibagi}\: \: x^{2}-1\: \: \textrm{bersisa}\: \: 2x-1\: ,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{dan}\: \: n\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad m=-1\: \: \textrm{dan}\: \: n=2&&\textrm{d}.\quad m=-1\: \: \textrm{dan}\: \: n=-2\\ \textrm{b}.\quad m=1\: \: \textrm{dan}\: \: n=-2&\textrm{c}.\quad m=1\: \: \textrm{dan}\: \: n=2&\textrm{e}.\quad m=-2\: \: \textrm{dan}\: \: n=1\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{c}\\\\ \textrm{Dengan menggunakan pembagian}\: \: \textbf{Horner-Kino}\: \: \textrm{akan diperoleh},\\\\ \begin{array}{rr|rrr|rrr} &&1&0&0&2m&-n\\ 1&&\ast &\ast &1&0&1&\\\cline{3-7} 0&&\ast &0&0&0&\ast &+\\\cline{3-7} \multicolumn{3}{r}{1}&0&1&2m&1-n\\\cline{6-7} \end{array}\quad \Rightarrow \begin{cases} \textrm{Suku banyak}: & f(x)=x^{4}+2mx-n \\ \textrm{Pembagai}: & p(x)=(x-1)(x+1)=x^{2}-1 \\ &: 1\: \: \textrm{dari}\: -\frac{-1}{1},\: \: \textrm{sedang}\: \: 0=-\left ( \frac{0}{1} \right )\\ \textrm{Hasil bagi}:&h(x)=x^{2}+1\\ \textrm{Sisa bagi}:&s(x)=2mx+(1-n)=2x-1 \end{cases} \\\\ \textrm{Sehingga},\\\\ 2m=2\Rightarrow m=1\\ 1-n=-1\Rightarrow n=2.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: f(x)=x^{4}-kx^{2}+5\: \: \textrm{habis dibagi}\: \: (x-1)\: \: , \textrm{maka}\: \: f(x)\: \: \textrm{juga habis dibagi oleh}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x+1&&\textrm{d}.\quad x+5\\ \textrm{b}.\quad 2x+1&\textrm{c}.\quad 3x+1&\textrm{e}.\quad 2x+5 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}f(x)&=x^{4}-kx^{2}+5\\ f(1)&=(1)^{4}-k(1)^{2}+5\\ 0&=1-k+5\\ k&=6\\ f(x)&=x^{4}-6x^{2}+5\\ &=(x^{2}-1)(x^{2}-5)\\ &=(x-1)(x+1)(x^{2}-5) \end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: (m-2)\: \: \textrm{adalah faktor dari}\: \: 2m^{3}+3tm+4\: \: , \textrm{maka nilai}\: \: t\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{10}{3}&&\textrm{d}.\quad -\displaystyle \frac{3}{10}\\ \textrm{b}.\quad \displaystyle \frac{1}{3}&\textrm{c}.\quad \displaystyle \frac{3}{10}&\textrm{e}.\quad -\displaystyle \frac{10}{3} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}f(m)&=2m^{3}+3tm+4\\ f(2)&=2(2)^{3}+3t(2)+4\\ 0&=16+6t+4\\ -6t&=20\\ t&=-\displaystyle \frac{10}{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{(KSM MA Kab/Kota 2015)Nilai terkecil}\: \: n\: \: \textrm{yang mengkin sehingga}\: \: n.(n+1).(n+2)\\\ & \textrm{habis dibagi 24 adalah}....\\ &\begin{array}{l}\\ \textrm{a}.\quad 1\\ \textrm{b}.\quad 2\\ \textrm{c}.\quad 3\\ \textrm{d}.\quad 4 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}k&=\displaystyle \frac{n.(n+1).(n+2)}{24}\\ &=\displaystyle \frac{n.(n+1).(n+2)}{2.(2+1).(2+2)}\\ \textrm{m}&\textrm{aka}\: \: n=2 \end{aligned}.

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