Contoh Soal Polinom (Bagian 8)

$\begin{array}{l}\\ 36.& (\mathbf{\text{Soal Seleksi OM Af-Sel 1983}})\\ & \text{Jika diketahui}{a}^3-a-1=0\text{maka}\\ & a^4+a^3-a^2-2a+1=....\\ & \begin{array}{l}\\ \text{a}.0& & \text{d}.1\\ \text{b}.a+1& & \text{e}.a^3+a+1\\ \text{c}.2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Lihat pembahasan no}.35\\ & \text{Cukup jelas }\end{array}\end{array}$ .

$\begin{array}{l}\\ 37.& \text{Tentukanlah suku banyak}f(x)\text{sedemikian}\\ & \text{sehingga}f(x)\text{terbagi oleh}{x}^2+1,\\ & \text{sedangkan}f(x)+1\text{terbagi oleh}{x}^3+x^2+1\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =(x^2+1).h_{1}x\\ f(x)+1& =(x^2+1).h_{1}x+1\\ \text{supaya}& f(x)+1\text{terbagi habis oleh}{x}^3+x^2+1,\\ & \text{maka akan ada bilangan bulat}k,(k\ne 0)\\ k& ={\displaystyle \frac{f(x)+1}{x^3+x^2+1}}\\ & ={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}}\\ k=1\Rightarrow & 1={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}\text{maka}{h}_{1}x=x}\\ \text{sehingga}& f(x)=x^3+x^2\\ \text{untuk ni}& \text{lai}k\text{yang lain, tak ditemukan}\end{array}\end{array}$.

$\begin{array}{l}\\ 38.& (\text{KSM 2015})\text{Diketahui}f(x)\text{adalah polinom}\\ & (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ & \text{dengan}{x}_1,x_2,x_3,x_4,\text{dan}{x}_5\text{adalah}\\ & \text{bilangan bulat berbeda}.\text{Jika}f(104)=2012,\\ & \text{maka nilai}{x}_1+x_2+x_3+x_4+x_5\text{sama dengan}....\\ & \text{a}.13\\ & \text{b}.14\\ & \text{c}.16\\ & \text{d}.17\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}:\\ f(x)& =(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ f(104)& =(104-x_1)(104-x_2)(104-x_3)(104-x_4)(104-x_5)=2012\\ & =2012=1\times 2\times 503\\ & =(-1)\times (1)\times (-2)\times (2)\times (503)\\ \text{maka}& \{\begin{array}{ll}(104-x_1)& =-2\Rightarrow x_1=106\\ (104-x_2)& =-1\Rightarrow x_2=105\\ (104-x_3)& =1\Rightarrow x_3=103\\ (104-x_4)& =2\Rightarrow x_4=102\\ (104-x_5)& =503\Rightarrow x_5=-399\end{array}\\ \text{sehin}& \text{gga},\\ & x_1+x_2+x_3+x_4+x_5=106+105+103+102+(-399)=17\end{array}\end{array}$.

$\begin{array}{l}\\ 39.& \text{Akar-akar persamaan}\\ & (x+1)(2x+1)(3x-1)(4x-1)+6x^4=0\\ & \text{adalah}{x}_1,x_2,x_3\text{dan}{x}_4.\text{Jika}\\ & x_1<x_2<x_3<x_4\text{dan}{x}_1+x_4=m\\ & \text{serta}{x}_2+x_3=n,\text{maka}mn=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{7}{30}}& & \text{d}.{\displaystyle \frac{2}{15}}\\ \text{b}.{\displaystyle \frac{7}{30}}& & \text{e}.-{\displaystyle \frac{4}{15}}\\ \text{c}.{\displaystyle \frac{4}{15}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & (x+1)(2x+1)(3x-1)(4x-1)+6x^4=0\\ & \iff (2x^2+3x+1)(12x^2-7x+1)+6x^4=0\\ & \iff 24x^4+22x^3-7x^2-4x+1+6x^4=0\\ & \iff 30x^4+22x^3-7x^2-4x+1+6x^4=0\\ & \iff (6x^2+2x+2)(5x^2+2x-1)=0\\ & \iff (6x^2+2x+2)=0\text{V}(5x^2+2x-1)=0\\ & \{\begin{array}{ll}{x}_1& ={\displaystyle \frac{-1-\sqrt{6}}{5}}\\ x_2& ={\displaystyle \frac{-1-\sqrt{7}}{6}}\end{array},\{\begin{array}{ll}{x}_3& ={\displaystyle \frac{-1+\sqrt{7}}{6}}\\ x_4& ={\displaystyle \frac{-1+\sqrt{6}}{5}}\end{array}\\ & \text{Dengan}m=x_1+x_4=-{\displaystyle \frac{2}{5},n=x_2+x_3=-{\displaystyle \frac{1}{3}}}\\ & \text{maka}mn=(-{\displaystyle \frac{2}{5}})(-{\displaystyle \frac{1}{3}})={\displaystyle \frac{2}{15}}\end{array}\end{array}$ 

$\begin{array}{l}\\ 40.& \text{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ & \text{adalah}{x}_1,x_2,x_3,...,x_{2017}\\ & \text{Tentukan nilai dari}\\ & {\displaystyle \frac{1}{1-x_1}+\frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{x^{2018}-1}{x-1}}& =x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ \text{perlu dii}& \text{ngat bahwa kondisi ini mensyaratkan}x\ne 1,\\ & \text{sehingga}\\ \\ x^{2018}-1& =0\\ x^{2018}& =1\\ x& =\pm 1,\text{pilih}x=-1\\ \text{maka}& \text{nilai dari}\\ {\displaystyle \frac{1}{1-x_1}+}& \frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+{\displaystyle \frac{1}{2}}}}}\\ & ={\displaystyle \frac{2017}{2}}\end{array}\end{array}$.