Contoh Soal Polinom (Bagian 2)

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

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

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

$.\begin{array}{rl}& \text{Sehingga},\\ & \bullet 2m=2\Rightarrow m=1\\ & \bullet 1-n=-1\Rightarrow n=2\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}f(x)=x^4-k{x}^2+5\text{habis dibagi}\\ & (x-1)\text{maka}f(x)\text{juga habis dibagi oleh}....\\ & \begin{array}{l}\\ \text{a}.x+1& & \text{d}.x+5\\ \text{b}.2x+1& \text{c}.3x+1& \text{e}.2x+5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =x^4-k{x}^2+5\\ f(1)& =(1{)}^4-k(1{)}^2+5\\ 0& =1-k+5\\ k& =6\\ f(x)& =x^4-6x^2+5\\ & =(x^2-1)(x^2-5)\\ & =(x-1)(x+1)(x^2-5)\end{array}\end{array}$.

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

$.\begin{array}{|l|}\hline \mathbf{\text{Catatan}}:\\ \\ \begin{array}{rl}& \text{Perhatikan uraian berikut}\\ & {\displaystyle \frac{x^3-12x+16}{(x-2)(x+4)}}\\ & ={\displaystyle \frac{x^3-12x+16}{x^2+2x-8}}\end{array}\\ \\ \begin{array}{lll}\mathbf{\text{pembagi}}& x-2\mathbf{\text{hasil}}& \mathbf{\text{bagi}}\\ x^2+2x-8& x^3-12x+16& \\ & x^3+2x^2-8x& -\\ & -2x^2-4x+16& \\ & -2x^2-4x+16& -\\ \mathbf{\text{Sisa}}& 0& (\mathbf{\text{habis}})\end{array}\\ \\ \begin{array}{rl}\therefore f(x)& =x^3-12x+16\\ & =(x-2{)}^2(x+4)\end{array}\\ \hline\end{array}$.

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

$.\begin{array}{|l|}\hline \mathbf{\text{Catatan}}:\\ \\ \begin{array}{rl}& \text{Perhatikan uraian berikut}\\ & {\displaystyle \frac{2x^3-9x^2+7x+6}{(x-2)(x-3)}}\\ & ={\displaystyle \frac{2x^3-9x^2+7x+6}{x^2-5x+6}}\end{array}\\ \\ \begin{array}{lll}\mathbf{\text{pembagi}}& 2x+1\mathbf{\text{hasil}}& \mathbf{\text{bagi}}\\ x^2-5x+6& 2x^3-9x^2+7x+6& \\ & 2x^3-10x^2+12x& -\\ & x^2-5x+6& \\ & x^2-5x+6& -\\ \mathbf{\text{Sisa}}& 0& (\mathbf{\text{habis}})\end{array}\\ \\ \begin{array}{rl}\therefore f(x)& =2x^3-9x^2+7x+6\\ & =(2x+1)(x-2)(x-3)\end{array}\\ \hline\end{array}$.