Contoh Soal Polinom (Bagian 2)

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

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: x^{4}+2mx-n\: \: \textrm{dibagi}\: \: x^{2}-1\\ &\textrm{bersisa}\: \: 2x-1\: ,\textrm{maka nilai}\: \: m\\ &\textrm{dan}\: \: n\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad m=-1\: \: \textrm{dan}\: \: n=2\\ \textrm{b}.\quad m=1\: \: \textrm{dan}\: \: n=-2\\ \textrm{c}.\quad \color{red}m=1\: \: \textrm{dan}\: \: \color{red}n=2\\ \textrm{d}.\quad m=-1\: \: \textrm{dan}\: \: n=-2\\ \textrm{e}.\quad m=-2\: \: \textrm{dan}\: \: n=1\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{dengan Horner-Kino didapatkan} \end{array}$

$.\qquad\begin{cases} \textrm{Suku banyak}: & f(x)=x^{4}+2mx-n \\ \textrm{Pembagai}: & p(x)=(x-1)(x+1)=x^{2}-1 \\ &: 1\: \: \textrm{dari}\: -\frac{-1}{1},\: \: \textrm{sedang}\: \: 0=-\left ( \frac{0}{1} \right )\\ \textrm{Hasil bagi}:&h(x)=x^{2}+1\\ \textrm{Sisa bagi}:&s(x)=2mx+(1-n)=2x-1 \end{cases}$

$.\qquad \begin{aligned}&\textrm{Sehingga},\\ &\bullet \quad 2m=2\Rightarrow m=\color{red}1\\ &\bullet \quad 1-n=-1\Rightarrow n=\color{red}2 \end{aligned}$

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

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: x^{3}-12x+k\: \: \textrm{habis dibagi oleh}\\  &(x-2)\: \: \textrm{maka polinom tersebut juga }\\ &\textrm{akan dibagi habis oleh}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x-1&&\textrm{d}.\quad x+2\\ \textrm{b}.\quad x-3&\textrm{c}.\quad x+1&\textrm{e}.\quad \color{red}x+4 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Misal}&\: \: f(x)=\color{blue}x^{3}-12x+k\\ \textrm{Saat}\: &f(2)=0\: \: (f(x)\: \: \textrm{habis dibagi}\: \: (x-2))\\ f(2)&=2^{3}-12.2+k=0\Leftrightarrow k=16\\ \textrm{Sehin}&\textrm{gga}\: \: f(x)=x^{3}-12x+16\\ \textrm{Deng}&\textrm{an teorema faktor, yang mungkin}\\ \textrm{adala}&\textrm{h}\: \: 16=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(-4)&=(-4)^{3}-12(-4)+16=0\\ \textrm{maka}&\: \: \color{red}x+4\: \: \color{black}\textrm{termasuk faktornya juga} \end{aligned} \end{array}$.

$.\: \qquad\begin{array}{|l|}\hline \textbf{Catatan}:\\\\ \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{3}-12x+16}{(x-2)\color{red}(x+4)}\\ &=\displaystyle \frac{x^{3}-12x+16}{x^{2}+2x-8}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x-2\qquad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline x^{2}+2x-8&x^{3}-12x+16&\\ &x^{3}+2x^{2}-8x&-\\\hline &-2x^{2}-4x+16&\\ &-2x^{2}-4x+16&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=x^{3}-12x+16\\ &=(x-2)^{2}\color{red}(x+4) \end{aligned}\\\hline \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: (x-2)\: \: \textrm{adalah faktor dari}\\  &f(x)=2x^{3}+ax^{2}+7x+6,\\ &\textrm{maka akar lainnya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x+3&&\textrm{d}.\quad 2x-3\\ \textrm{b}.\quad \color{red}x-3&\textrm{c}.\quad x-1&\textrm{e}.\quad 2x+3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Misal}&\: \: f(x)=\color{blue}2x^{3}+ax^{2}+7x+6\\ \textrm{Saat}\: &f(2)=0\: \: (f(x)\: \: \textrm{habis dibagi}\: \: (x-2))\\ f(2)&=2.2^{3}+a.2^{2}+7.2+6=0\Leftrightarrow a=-9\\ \textrm{Sehin}&\textrm{gga}\: \: f(x)=2x^{3}-9x^{2}+7x+6\\ \textrm{Deng}&\textrm{an teorema faktor, yang mungkin}\\ \textrm{adala}&\textrm{h}\: \: \displaystyle \frac{6}{2}=\pm 1,\pm 2,\pm 3\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(3)&=2(3)^{3}-9(3)+7.3+6=0\\ \textrm{maka}&\: \: \color{red}x-3\: \: \color{black}\textrm{termasuk faktornya juga} \end{aligned} \end{array}$.

$.\: \qquad\begin{array}{|l|}\hline \textbf{Catatan}:\\\\ \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{2x^{3}-9x^{2}+7x+6}{(x-2)\color{red}(x-3)}\\ &=\displaystyle \frac{2x^{3}-9x^{2}+7x+6}{x^{2}-5x+6}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}2x+1\qquad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline x^{2}-5x+6&2x^{3}-9x^{2}+7x+6&\\ &2x^{3}-10x^{2}+12x&-\\\hline &\: \: \: \: \: \qquad x^{2}-5x+6&\\ &\: \: \: \: \: \qquad x^{2}-5x+6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=2x^{3}-9x^{2}+7x+6\\ &=\color{red}(2x+1)\color{black}(x-2)\color{red}(x-3) \end{aligned}\\\hline \end{array}$.