Contoh Soal Polinom (Bagian 8)

$\begin{array}{ll}\\ 36.&(\textbf{Soal Seleksi OM Af-Sel 1983})\\ &\textrm{Jika diketahui}\: \: a^{3}-a-1=0\: \: \textrm{maka}\\ & a^{4}+a^{3}-a^{2}-2a+1\: =\:  ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0&&\textrm{d}.\quad 1\\ \textrm{b}.\quad  a+1&\qquad&\textrm{e}.\quad a^{3}+a+1\\ \textrm{c}.\quad  \color{red}2\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Lihat pembahasan no}.35\\ &\textrm{Cukup jelas }   \end{aligned}   \end{array}$ .

$\begin{array}{ll}\\ 37.&\textrm{Tentukanlah suku banyak}\: \: f(x)\: \: \textrm{sedemikian}\\ &\textrm{sehingga}\: \: f(x)\: \: \textrm{terbagi oleh}\: \: x^{2}+1,\\ &\textrm{sedangkan}\: \: f(x)+1\: \: \textrm{terbagi oleh}\: \: x^{3}+x^{2}+1\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\left ( x^{2}+1 \right ).h_{1}x\\ f(x)+1&=\left ( x^{2}+1 \right ).h_{1}x+1\\ \textrm{supaya}\: \: \: &f(x)+1\: \: \textrm{terbagi habis oleh}\: \: x^{3}+x^{2}+1 ,\\ & \textrm{maka akan ada bilangan bulat}\: \: k,\: \: \left ( k\neq 0 \right )\\ k&=\displaystyle \frac{f(x)+1}{x^{3}+x^{2}+1}\\ &=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\\ k=1\Rightarrow &1=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\: \: \textrm{maka}\: \: h_{1}x=x\\ \textrm{sehingga}&\: \: f(x)=\color{red}x^{3}+x^{2}\\ \textrm{untuk ni}&\textrm{lai}\: \: k\: \: \textrm{yang lain, tak ditemukan} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\color{blue}(\textrm{KSM 2015})\color{black}\textrm{Diketahui}\: \: f(x)\: \: \textrm{adalah polinom}\\ & (x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})(x-x_{5})\\ &\textrm{dengan}\: \: \: x_{1},\: x_{2},\: x_{3},\: x_{4},\: \: \textrm{dan}\: \: x_{5}\: \: \textrm{adalah}\\ &\textrm{bilangan bulat berbeda}.\: \textrm{Jika}\: \: f(104)=2012,\\ &\textrm{maka nilai} \: \: \: x_{1}+ x_{2}+ x_{3}+ x_{4}+x_{5}\: \: \textrm{sama dengan}....\\ &\textrm{a}.\quad 13\\ &\textrm{b}.\quad 14\\ &\textrm{c}.\quad 16\\ &\textrm{d}.\quad \color{red}17\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}:\\ f(x)&=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})(x-x_{5})\\ f(104)&=(104-x_{1})(104-x_{2})(104-x_{3})(104-x_{4})(104-x_{5})=2012\\ &=2012=1\times 2\times 503\\ &=(-1)\times (1)\times (-2)\times (2)\times (503)\\ \textrm{maka}&\: \: \begin{cases} (104-x_{1}) &=-2\Rightarrow x_{1}=106 \\ (104-x_{2}) &=-1\Rightarrow x_{2}=105 \\ (104-x_{3}) &=1\Rightarrow x_{3}=103 \\ (104-x_{4}) &=2\Rightarrow x_{4}=102 \\ (104-x_{5}) &=503\Rightarrow x_{5}=-399 \\ \end{cases}\\ \textrm{sehin}&\textrm{gga},\\ &x_{1}+ x_{2}+ x_{3}+ x_{4}+x_{5}=106+105+103+102+(-399)=\color{red}17 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Akar-akar persamaan}\\ &(x+1)(2x+1)(3x-1)(4x-1)+6x^{4}=0\\ &\textrm{adalah}\: \:  x_{1},x_{2},x_{3}\: \: \textrm{dan}\: \: x_{4}\: .\: \textrm{Jika}\\ &x_{1}<x_{2}<x_{3}< x_{4}\: \: \textrm{dan}\: \: x_{1}+x_{4}=m\\ &\textrm{serta}\: \: x_{2}+x_{3}=n,\: \textrm{maka}\: \: mn=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{7}{30}&&\textrm{d}.\quad \color{red}\displaystyle \frac{2}{15}\\ \textrm{b}.\quad  \displaystyle \frac{7}{30}&\qquad&\textrm{e}.\quad -\displaystyle \frac{4}{15}\\ \textrm{c}.\quad  \displaystyle \frac{4}{15}\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &(x+1)(2x+1)(3x-1)(4x-1)+6x^{4}=0\\ &\Leftrightarrow (2x^{2}+3x+1)(12x^{2}-7x+1)+6x^{4}=0\\ &\Leftrightarrow 24x^{4}+22x^{3}-7x^{2}-4x+1+6x^{4}=0\\ &\Leftrightarrow 30x^{4}+22x^{3}-7x^{2}-4x+1+6x^{4}=0\\ &\Leftrightarrow (6x^{2}+2x+2)(5x^{2}+2x-1)=0\\ &\Leftrightarrow (6x^{2}+2x+2)=0\: \textrm{V}\: (5x^{2}+2x-1)=0\\ &\begin{cases} x_{1} &=\displaystyle \frac{-1-\sqrt{6}}{5} \\  x_{2} &=\displaystyle \frac{-1-\sqrt{7}}{6} \end{cases},\: \begin{cases} x_{3} &=\displaystyle \frac{-1+\sqrt{7}}{6} \\  x_{4} &=\displaystyle \frac{-1+\sqrt{6}}{5}  \end{cases}\\ &\textrm{Dengan}\: \: m=x_{1}+x_{4}=-\displaystyle \frac{2}{5},\: n=x_{2}+x_{3}=-\displaystyle \frac{1}{3}\\ &\textrm{maka}\: \: mn=\left ( -\displaystyle \frac{2}{5} \right )\left ( -\displaystyle \frac{1}{3} \right )=\color{red}\displaystyle \frac{2}{15}  \end{aligned}    \end{array}$ 

$\begin{array}{ll}\\ 40.&\textrm{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^{2}+x+1=0\\ & \textrm{adalah}\: \: x_{1},\: x_{2},\: x_{3},...,x_{2017}\\ &\textrm{Tentukan nilai dari}\\ & \displaystyle \frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\frac{1}{1-x_{3}}+...+\displaystyle \frac{1}{1-x_{2017}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{x^{2018}-1}{x-1}&=x^{2017}+x^{2016}+x^{2015}+...+x^{2}+x+1=0\\ \textrm{perlu dii}&\textrm{ngat bahwa kondisi ini mensyaratkan}\: \: x\neq 1,\\ & \textrm{sehingga}\\\\ x^{2018}-1&=0\\ x^{2018}&=1\\ x&=\pm 1,\: \: \: \: \textrm{pilih}\: \: x=-1\\ \textrm{maka}\: \quad &\textrm{nilai dari}\\ \displaystyle \frac{1}{1-x_{1}}+&\frac{1}{1-x_{2}}+\frac{1}{1-x_{3}}+...+\displaystyle \frac{1}{1-x_{2017}}\\ &=\underset{\textrm{sebanyak 2017}}{\underbrace{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}\\ &=\underset{\textrm{sebanyak 2017}}{\underbrace{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+\displaystyle \frac{1}{2}}} \\ &=\color{red}\displaystyle \frac{2017}{2} \end{aligned} \end{array}$.