Metode Horner-Kino (Lanjutan Materi Operasi Polinom)

 TAMBAHAN

Pembagian Horner - Kino

Perhatikanlah bagan berikut

Sebagai tambahan penjelasan dari bagan di atas adalah

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Dengan metode Horner, tentukanlah}\\ & \text{nilai suku banyak berikut ini}!\\ & \text{a})4x^4-7x^3+8x^2-2x+3\text{jika}x=2\\ & \text{b})2x^5+3x^3-x+1\text{jika}x=-3\\ & \text{c})2x^3+x^2-2x+3\text{jika}x={\displaystyle \frac{1}{3}}\\ \\ & \text{Jawab}:\\ & \text{Diketahui bahwa}\\ & f(x)=4x^4-7x^3+8x^2-2x+3\\ & \mathbf{\text{Cara biasa (Substitusi)}}\\ & \begin{array}{rl}f(2)& =4(2{)}^4-7(2{)}^3+8(2{)}^2-2(2)+3\\ & =64-56+32-4+3\\ & =39\\ \text{Seba}& \text{gai catatan bahwa}:\\ & \text{Polinom}f(x)\text{tersebut di atas }\\ & \text{jika dibagi}(x-2)\text{bersisa 39}\end{array}\\ & \mathbf{\text{Cara Horner}}\\ & \end{array}$

$\begin{array}{l}\\ 2.& \text{Hitunglah nilai}a,b,c,\text{dan}d,\text{jika}\\ & \text{a})-3x+4\equiv a(x-7)-b(2x-3)\\ & \text{b})a(x-1{)}^2-b(x+4)\equiv 2x^2-5x-7\\ & \text{c})3x^2+2x-5\equiv (ax+1)(x+b)-c(x+1)+2(ab-c)\\ & \text{d})x^4-8x^3+15x-20\equiv x^4+a{x}^3+(a+b)x^2+(2b-c)x+d\\ & \text{e}){\displaystyle \frac{a}{x-1}+\frac{b}{x+3}\equiv {\displaystyle \frac{8}{x^2+2x-3}}}\\ & \text{f}){\displaystyle \frac{a}{x-1}+\frac{b}{x-4}\equiv {\displaystyle \frac{3}{x-1}+\frac{20}{x-4}+\frac{x+17}{x^2-5x+4}}}\\ & \text{g}){\displaystyle \frac{5x-4}{x^2-1}\equiv {\displaystyle \frac{a}{x-1}+\frac{b}{x+1}-\frac{3}{x^2-1}}}\\ & \text{h}){\displaystyle \frac{2x^2+x+2}{x^3-1}\equiv {\displaystyle \frac{a}{x-1}+\frac{bx+c}{x^2+x+1}}}\\ & \text{i}){\displaystyle \frac{3x^2+2x-5}{x^2+5x+6}\equiv {\displaystyle \frac{a(x-3)}{x+3}+\frac{b(x-5)}{x+2}+\frac{4c}{(x+2)(x+3)}}}\\ & \text{j})x^3+a{x}^2+bx+c=0\text{dengan akar-akar}{x}_1=x_2=-1\text{dan}{x}_3=-3\\ & \text{k})x^3+a{x}^2+bx+c=0\text{dengan akar-akar}1,2,\text{dan}3\end{array}$

$.\begin{array}{rl}\text{Yang diba}& \text{has hanya no. 6 d}\\ x^4-8x^3+& 15x-20\\ \equiv x^4+a{x}^3& +(a+b)x^2+(2b-c)x+d\\ \text{koefisien}{x}^4& :1=1\\ \text{koefisien}{x}^3& :-8=a,\text{maka}a=-8\\ \text{koefisien}{x}^2& :0=a+b,\text{maka}b=-a=-(-8)=8\\ \text{koefisien}{x}^1& :15=2b-c,\text{maka}c=2b-15=2(8)-15=1\\ \text{koefisien}{x}^0& :-20=d,\text{maka}d=-20\end{array}$

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

$.\begin{array}{rl}& \text{Untuk pembahasan no. 3 i}\end{array}$

$.\begin{array}{rl}& \begin{array}{|ll|}\hline \text{Pembagi}& \begin{array}{rl}& \text{Sisa}\\ & s(x)={\displaystyle \frac{7}{2}}\end{array}\\ \begin{array}{rl}& 2x-1=2(x-\frac{1}{2})\end{array}& \begin{array}{rl}& \text{Hasil bagi}\\ & {\displaystyle \frac{h(x)}{2}=\frac{2x^2+8x-1}{2}=x^2+4x-\frac{1}{2}}\end{array}\\ \hline\end{array}\end{array}$

$.\begin{array}{rl}& \text{Dan untuk pembahasan no. 3 m}\end{array}$

$.\begin{array}{|ll|}\hline \text{Pembagi}& \begin{array}{rl}& \text{Sisa}\\ & s_2(x-p)+s_1\\ & 1(x-1)+0=x-1\end{array}\\ \begin{array}{rl}(x-p)(x-q)& =(x-1)(x-1)\\ & =(x-1{)}^2\end{array}& \begin{array}{rl}& \text{Hasil bagi}\\ & 2x^2+4x+3\end{array}\\ \hline\end{array}$

$.\begin{array}{rl}& \text{Coba bandingkan dengan cara Horner-Kino berikut}\end{array}$

$.\{\begin{array}{ll}\text{Suku banyak}:& f(x)=2x^4-3x^2-x+2\\ \text{Pembagai}:& p(x)=x^2-2x+1\\ & :-1\text{dari}-\frac{1}{1},\text{sedang}2=-\left(\frac{-2}{1}\right)\\ \text{Hasil bagi}:& h(x)=2x^2+4x+3\\ \text{Sisa bagi}:& s(x)=x-1\end{array}$.

$$\begin{gathered} .\text{Sehingga}, \\ 2x^4-3x^2-x+2=(x^2-2x+1)(2x^2+4x+3)+x-1 \end{gathered}$$

$\begin{array}{l}\\ 4.& \text{Jika diketahui akar-akar persamaan}{x}^2+4x-5=0\\ & \text{juga akar-akar untuk persamaan}2x^3+9x^2-6x-5=0,\\ & \text{maka akar ketiga untuk persamaan yang kedua adalah}...\\ \\ & \text{Jawab}:\end{array}$

$.\{\begin{array}{ll}\text{Suku banyak}:& f(x)=2x^3+9x^2-6x-5\\ \text{Pembagai}:& p(x)=x^2+4x-5\\ & :5\text{dari}-\left(\frac{-5}{1}\right),\text{sedang}-4=\left(\frac{4}{1}\right)\\ \text{Hasil bagi}:& h(x)=2x+1\\ \text{Sisa bagi}:& s(x)=0\end{array}$

$.\begin{array}{rl}& \text{Sehingga}\\ & 2x^3+9x^2-6x-5=(x^2+4x-5)(2x+1)\\ & \text{Jadi, akar yang lain (yang ketiga) adalah}\\ & (2x+1)\Rightarrow x=-{\displaystyle \frac{1}{2}}\end{array}$

$\text{LATIHAN SOAL}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah hasil bagi dan sisanya}!\\ & \begin{array}{l}\\ \text{a})(x^7+3x^5+1):(x^2-1)\\ \text{b})(x^4-3x^3-5x^2+x-6):(x^2-x-2)\\ \text{c})(2x^3+x^2-4x+5):(x^2+x+1)\\ \text{d})(2x^4+x^3-3x+6):(x^2+x+2)\\ \text{e})(x^4-3x^2+7x-4):(x^2-2x-1)\\ \text{f})(3x^3+4x-8):(3x^2+x+2)\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}a\text{dan}b\text{bilangan bulat yang menyebabkan}\\ & x^2-x-1\text{merupakan faktor dari}a{x}^3+b{x}^2+1,\\ & \text{maka harga}b\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2& & & \text{d}.& 1\\ \text{b}.& -1& \text{c}.& 0& \text{e}.& 2\end{array}\\ & (\text{AHSME 1988})\end{array}$.

$\text{Pembagian Istimewa}$

Aturan pembagian istimewa adalah

$\begin{array}{rl}1.& {\displaystyle \frac{x^n-a^n}{x-a}=x^{n-1}{a}^0+x^{n-2}{a}^1+\cdots +x^1{a}^{n-2}+x^0{a}^{n-1}}\\ & ={\displaystyle \sum _{k=1}^n{x}^{n-k}{a}^{k-1}}\\ & \text{dengan suku ke}-k\text{hasil bagi}=x^{n-k}{a}^{k-1}\\ 2.& {\displaystyle \frac{x^{2n}-a^{2n}}{x+a}=x^{2n-1}{a}^0-x^{2n-2}{a}^1+\cdots +x^1{a}^{2n-2}-x^0{a}^{2n-1}}\\ & ={\displaystyle \sum _{k=1}^{2n}(-1{)}^{k+1}{x}^{2n-k}{a}^{k-1}}\\ & \text{dengan suku ke}-k\text{hasil bagi}=(-1{)}^{k+1}{x}^{2n-k}{a}^{k-1}\\ 3.& {\displaystyle \frac{x^{2n+1}+a^{2n+1}}{x+a}=x^{2n}{a}^0-x^{2n-1}{a}^1+\cdots -x^1{a}^{2n-1}+x^0{a}^{2n}}\\ & ={\displaystyle \sum _{k=1}^{2n+1}(-1{)}^{k+1}{x}^{2n+1-k}{a}^{k-1}}\\ & \text{dengan suku ke}-k\text{hasil bagi}=(-1{)}^{k+1}{x}^{2n+1-k}{a}^{k-1}\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah hasil bagi polinom}\\ & \text{untuk tiap pembagian istimewa berikut}\\ & \text{a}.(x^3-a^3):(x-a)\\ & \text{b}.(x^4-a^4):(x+a)\\ & \text{c}.(x^5+a^5):(x+a)\\ \\ & \text{Jawab}:\\ & \text{a}.{\displaystyle \frac{(x^3-a^3)}{(x-a)}=x^2+xa+a^2....(\text{rumus}1)}\\ & \text{b}.{\displaystyle \frac{(x^4-a^4)}{(x+a)}=x^3-x^{2}a+x{a}^2-a^3....(\text{rumus}2)}\\ & \text{c}.{\displaystyle \frac{(x^5+a^5)}{(x+a)}=x^4-x^{3}a+x^2{a}^2-x{a}^3+a^4....(\text{rumus}3)}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah hasil bagi polinom}\\ & \text{untuk tiap pembagian istimewa berikut}\\ & \text{a}.(m^8-n^8):(m+n)\\ & \text{b}.(x^{10}-y^{10}):(x+y)\\ \\ & \text{Jawab}:\\ & \text{a}.{\displaystyle \frac{(m^8-n^8)}{(m+n)}=m^7-m^{6}n+m^5{n}^2-\cdots +m{n}^6-n^7}\\ & \text{b}.{\displaystyle \frac{(x^{10}-y^{10})}{(x+y)}=x^9-x^{8}y+x^7{y}^2-\cdots +x{y}^8-y^9}\end{array}$