Contoh Soal 3 Polinom

$\begin{array}{ll}\\ 11.&\textrm{Jika polinom}\: \: f(x)\: \: \textrm{dibagi oleh}\\ &(x-a)(x-b)\: \: \textrm{dan}\: \: a\neq b\: ,\: \textrm{maka}\\ &\textrm{sisa pembagiannya adalah}\: ....\\ &\begin{array}{lllllll}\\ &\textrm{a}.\quad \displaystyle \displaystyle \frac{x-a}{a-b}f(a)+\frac{x-a}{b-a}f(b)\\\\ &\textrm{b}.\quad \displaystyle \displaystyle \frac{x-a}{a-b}f(b)+\frac{x-a}{b-a}f(a)\\\\ &\textrm{c}.\quad \displaystyle \displaystyle \color{red}\frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)\\\\ &\textrm{d}.\quad \displaystyle \displaystyle \frac{x-b}{a-b}f(b)+\frac{x-a}{b-a}f(a)\\\\ &\textrm{e}.\quad \displaystyle \displaystyle \frac{x-a}{b-a}f(b)+\frac{x-a}{b-a}f(a)\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Misal sisa pembagiannya}:\: \color{red}s(x)=px+q\\ &\textrm{Saat}\: \: f(x)\: \: \textrm{dibagi}\: \: (x-a)(x-b)\: \: \textrm{berarti}\\ &\bullet \quad x=a\Rightarrow s(a)=f(a)=ap+q\: ....(1)\\ &\bullet \quad x=b\Rightarrow s(b)=f(b)=bp+q\: ......(2)\\ &\textrm{Persamaan}\: \: (1)\: \: \textrm{dan}\: \: (2)\: \: \textrm{dieliminasi}\\ &\color{blue}\begin{array}{llllllll}\\ ap&+&q&=&f(a)\\ bp&+&q&=&f(b)&-\\\hline ap&-&bp&=&f(a)-f(b)\\ &&p&=&\color{purple}\displaystyle \frac{f(a)-f(b)}{a-b}& \end{array}\\ &\textrm{Dari persamaan}\: \: (1),\\ &f(a)=ap+q\\ &f(a)=a\left ( \displaystyle \frac{f(a)-f(b)}{a-b} \right )+q\\ &q=a\left ( \displaystyle \frac{f(a)-f(b)}{a-b} \right )+f(a)\\ &q=a\left ( \displaystyle \frac{f(a)-f(b)}{a-b} \right )+f(a)\left ( \displaystyle \frac{a-b}{a-b} \right )\\ &q=\displaystyle \frac{-bf(a)-af(b)}{a-b}\\ &\textrm{Sehingga}\\ &s(x)=px+q\\ &\qquad =\left ( \displaystyle \frac{f(a)-f(b)}{a-b} \right )x+\left ( \displaystyle \frac{-bf(a)-af(b)}{a-b} \right )\\ &\qquad =\displaystyle \frac{f(a)x-f(b)x-bf(a)+af(b)}{a-b}\\ &\qquad =\displaystyle \frac{(x-b)f(a)+(a-x)f(b)}{a-b}\\ &\qquad =\displaystyle \frac{x-b}{a-b}f(a)+\frac{a-x}{a-b}f(b)\\ &\qquad =\color{red}\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Diketahui}\: \: f(x)\: \: \textrm{dibagi oleh}\: \: x-2\: \: \textrm{bersisa 5},\\ &\textrm{dan dibagi}\: \: x-3\: \: \textrm{bersisa 7. Jia}\: \: f(x)\: \: \\ &\textrm{dibagi oleh}\: \: x^{2}-5x+6\: \: \textrm{akan memiliki sisa}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x-2&&\textrm{d}.\quad \color{red}\displaystyle 2x+1\\ \textrm{b}.\quad \displaystyle 2x-4&\textrm{c}.\quad \displaystyle x+2&\textrm{e}.\quad 2x+3 \end{array}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}f(x)&=(x-2).h(x)+5\\ f(x)&=(x-3).h(x)+7\\ f(x)&=(x^{2}-5x+6).H(x)+s(x)\\ f(x)&=(x-2)(x-3).H(x)+px+q\\ f(2)&=(2-2)(2-3).H(x)+2p+q=5\\ &\Rightarrow \color{blue}0+2p+q=5\: \color{black}.................(1)\\ f(3)&=(3-2)(3-3).H(x)+3p+q=7\\ &\Rightarrow \color{blue}0+3p+q=7\: \color{black}.................(2)\\ \textrm{Dari}&\: \textrm{persamaan}\: \: (1)\: \: \textrm{dan}\: \: (2)\\ \color{red}\textrm{saat}\: &\color{red}\textrm{persamaan (1) dikurangi persamaan (2)}\\ &\qquad -p=-2\\ &\qquad\: \: \: \: \: \: p=2\\ &\textrm{maka}, \: \: \: q=1\\ &\textrm{Sehingga},\: \: \\ &s(x)=px+q=\color{red}2x+1\end{aligned}\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&f(x)\: \: \textrm{dibagi}\: \: (x-2)\: \: \textrm{sisa}\: \: 5\: \Rightarrow f(2)=5\\ &f(x)\: \: \textrm{dibagi}\: \: (x-3)\: \: \textrm{sisa}\: \: 7\: \Rightarrow f(3)=7\\ &\textrm{maka},\\ &s(x)=\color{red}\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)\\ &\qquad =\color{red}\displaystyle \frac{x-3}{2-3}\color{black}(5)\color{red}+\frac{x-2}{3-2}\color{black}(7)\\ &\qquad =\displaystyle \frac{5x-15}{-1}+\frac{7x-14}{1}\\ &\qquad =15-5x+7x-14\\ &\qquad =\color{red}2x+1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Polinom}\: \: f(x)\: \: \textrm{dibagi oleh}\: \: (2x-4)\: \: \textrm{bersisa 6},\\ &\textrm{dibagi oleh}\: \: (x+4)\: \: \textrm{bersisa 24}.\\ &\textrm{Dan polinom}\: \: g(x)\: \: \textrm{dibagi oleh}\: \: (2x-4)\: \: \textrm{bersisa 5},\\ & \textrm{dibagi oleh}\: \: (x+4)\: \: \textrm{bersisa 2}.\\ &\textrm{Jika}\: \: h(x)=f(x).g(x),\: \: \textrm{maka}\: \: h(x)\\ &\textrm{dibagi}\: \: (2x^{2}+4x-16)\: \: \textrm{akan sisa}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -3x+24&&\textrm{d}.\quad -6x+36\\ \textrm{b}.\quad \color{red}-3x+36&\textrm{c}.\quad 6x+24&\textrm{e}.\quad 12x+3 \end{array}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textrm{Langkah pertama}\\ &\begin{aligned}f(x)&=(2x-4).h(x)_{1}+6\\ f(x)&=(x+4).h(x)_{2}+24\\ f(x)&=(2x-4)(x+4).H_{1}(x)+p_{1}x+q_{1}\\ &\textrm{Gunakanlah cara sebagai mana}\\ &\textrm{contoh soal No. 12 di atas yang}\\ \color{magenta}\textrm{Alte}&\color{magenta}\textrm{natif 2}\\ \textrm{mak}&\textrm{a}\quad p_{1}x+q_{1}=-3x+12 \end{aligned} \\ &\color{blue}\textrm{Langkah kedua}\\ &\begin{aligned}g(x)&=(2x-4).h(x)_{3}+5\\ g(x)&=(x+4).h(x)_{4}+2\\ g(x)&=(2x-4)(x+4).H_{2}(x)+p_{2}x+q_{2}\\ &\textrm{Gunakanlah cara sebagai mana}\\ &\textrm{contoh soal No. 12 di atas yang}\\ \color{magenta}\textrm{Alte}&\color{magenta}\textrm{natif 2}\\ \textrm{mak}&\textrm{a}\quad p_{2}x+q_{2}=\displaystyle \frac{1}{2}x+4 \end{aligned} \\ &\color{blue}\textrm{Langkah ketiga}\\ &\begin{aligned}&h(x)=\color{red}f(x)\times g(x)\\ &=\left ( (2x-4)(x+4)H_{1}(x)+(-3x+12) \right )\\ &\qquad\qquad\qquad \times \left ( (2x-4)(x+4)H_{2}(x)+\displaystyle \frac{1}{2}x+4 \right )\\ &\textrm{maka}\\ &\bullet \quad h(2)=\left ( 0+(-3.2+12) \right )\left ( 0+\displaystyle \frac{1}{2}.2+4 \right )=6.5=30\\ &\bullet \quad h(-4)=\left ( 0+(-3.-4+12) \right )\left ( 0+\displaystyle \frac{1}{2}.-4+4 \right )=24.2=48\\ &\textrm{Dengan pembagi}\: \: 2x^{2}+x-16,\: \textrm{maka sisanya}:\: s_{3}(x)=p_{3}x+q_{3}\\ &\textrm{saat}\: \: x=2\qquad \Rightarrow 2p+q=30\\ &\textrm{saat}\: \: x=-4\: \: \Rightarrow -4p+q=48\\ &\textrm{selanjutnya dengan eliminasi-substitusi diperoleh}\: \: p=-3,\: q=36\\ &\textrm{sehingga}\: \: s(x)=px+q=\color{red}-3x+36 \end{aligned} \end{array}$

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

$\begin{array}{ll}\\ 15.&\textrm{Tentukanlah suku banyak}\: \: f(x)\: \: \textrm{sedemikian}\\ &\textrm{sehingga}\: \: f(x)\: \: \textrm{terbagi oleh}\: \: x^{2}+1,\\ &\textrm{sedangkan}\: \: f(x)+1\: \: \textrm{terbagi oleh}\: \: x^{3}+x^{2}+1\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\left ( x^{2}+1 \right ).h_{1}x\\ f(x)+1&=\left ( x^{2}+1 \right ).h_{1}x+1\\ \textrm{supaya}\: \: \: &f(x)+1\: \: \textrm{terbagi habis oleh}\: \: x^{3}+x^{2}+1 ,\\ & \textrm{maka akan ada bilangan bulat}\: \: k,\: \: \left ( k\neq 0 \right )\\ k&=\displaystyle \frac{f(x)+1}{x^{3}+x^{2}+1}\\ &=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\\ k=1\Rightarrow &1=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\: \: \textrm{maka}\: \: h_{1}x=x\\ \textrm{sehingga}&\: \: f(x)=\color{red}x^{3}+x^{2}\\ \textrm{untuk ni}&\textrm{lai}\: \: k\: \: \textrm{yang lain, tak ditemukan} \end{aligned} \end{array}$

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