Contoh Soal Polinom (Bagian 4)

$\begin{array}{l}\\ 16.& \text{Jika}(m-2)\text{adalah faktor dari}2m^3+3tm+4,\\ & \text{maka nilai}t\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{10}{3}}& & \text{d}.-{\displaystyle \frac{3}{10}}\\ \\ \text{b}.{\displaystyle \frac{1}{3}}& \text{c}.{\displaystyle \frac{3}{10}}& \text{e}.-{\displaystyle \frac{10}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(m)& =2m^3+3tm+4\\ f(2)& =2(2{)}^3+3t(2)+4\\ 0& =16+6t+4\\ -6t& =20\\ t& =-{\displaystyle \frac{10}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{(KSM MA Kab/Kota 2015)Nilai terkecil}n\\ & \text{yang mengkin sehingga}n.(n+1).(n+2)\\ & \text{habis dibagi 24 adalah}....\\ & \begin{array}{c}\\ \text{a}.1\\ \text{b}.2\\ \text{c}.3\\ \text{d}.4\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}k& ={\displaystyle \frac{n.(n+1).(n+2)}{24}}\\ & ={\displaystyle \frac{n.(n+1).(n+2)}{2.(2+1).(2+2)}}\\ & \text{maka}n=2\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Jika polinom}f(x)\text{dibagi oleh}\\ & (x-a)(x-b)\text{dan}a\ne b,\text{maka}\\ & \text{sisa pembagiannya adalah}....\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle {\displaystyle \frac{x-a}{a-b}f(a)+\frac{x-a}{b-a}f(b)}}\\ \\ & \text{b}.{\displaystyle {\displaystyle \frac{x-a}{a-b}f(b)+\frac{x-a}{b-a}f(a)}}\\ \\ & \text{c}.{\displaystyle {\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}}\\ \\ & \text{d}.{\displaystyle {\displaystyle \frac{x-b}{a-b}f(b)+\frac{x-a}{b-a}f(a)}}\\ \\ & \text{e}.{\displaystyle {\displaystyle \frac{x-a}{b-a}f(b)+\frac{x-a}{b-a}f(a)}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misal sisa pembagiannya}:s(x)=px+q\\ & \text{Saat}f(x)\text{dibagi}(x-a)(x-b)\text{berarti}\\ & \bullet x=a\Rightarrow s(a)=f(a)=ap+q....(1)\\ & \bullet x=b\Rightarrow s(b)=f(b)=bp+q......(2)\\ & \text{Persamaan}(1)\text{dan}(2)\text{dieliminasi}\\ & \begin{array}{l}\\ ap& +& q& =& f(a)\\ bp& +& q& =& f(b)& -\\ ap& -& bp& =& f(a)-f(b)\\ & & p& =& {\displaystyle \frac{f(a)-f(b)}{a-b}}& \end{array}\\ & \text{Dari persamaan}(1),\\ & f(a)=ap+q\\ & f(a)=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+q\\ & q=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+f(a)\\ & q=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+f(a)\left({\displaystyle \frac{a-b}{a-b}}\right)\\ & q={\displaystyle \frac{-bf(a)-af(b)}{a-b}}\\ & \text{Sehingga}\\ & s(x)=px+q\\ & =\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)x+\left({\displaystyle \frac{-bf(a)-af(b)}{a-b}}\right)\\ & ={\displaystyle \frac{f(a)x-f(b)x-bf(a)+af(b)}{a-b}}\\ & ={\displaystyle \frac{(x-b)f(a)+(a-x)f(b)}{a-b}}\\ & ={\displaystyle \frac{x-b}{a-b}f(a)+\frac{a-x}{a-b}f(b)}\\ & ={\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}\end{array}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Diketahui}f(x)\text{dibagi oleh}x-2\text{bersisa 5},\\ & \text{dan dibagi}x-3\text{bersisa 7. Jia}f(x)\\ & \text{dibagi oleh}{x}^2-5x+6\text{akan memiliki sisa}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x-2}& & \text{d}.{\displaystyle 2x+1}\\ \text{b}.{\displaystyle 2x-4}& \text{c}.{\displaystyle x+2}& \text{e}.2x+3\end{array}\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}f(x)& =(x-2).h(x)+5\\ f(x)& =(x-3).h(x)+7\\ f(x)& =(x^2-5x+6).H(x)+s(x)\\ f(x)& =(x-2)(x-3).H(x)+px+q\\ f(2)& =(2-2)(2-3).H(x)+2p+q=5\\ & \Rightarrow 0+2p+q=5.................(1)\\ f(3)& =(3-2)(3-3).H(x)+3p+q=7\\ & \Rightarrow 0+3p+q=7.................(2)\\ \text{Dari}& \text{persamaan}(1)\text{dan}(2)\\ \text{saat}& \text{persamaan (1) dikurangi persamaan (2)}\\ & -p=-2\\ & p=2\\ & \text{maka},q=1\\ & \text{Sehingga},\\ & s(x)=px+q=2x+1\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& f(x)\text{dibagi}(x-2)\text{sisa}5\Rightarrow f(2)=5\\ & f(x)\text{dibagi}(x-3)\text{sisa}7\Rightarrow f(3)=7\\ & \text{maka},\\ & s(x)={\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}\\ & ={\displaystyle \frac{x-3}{2-3}(5)+\frac{x-2}{3-2}(7)}\\ & ={\displaystyle \frac{5x-15}{-1}+\frac{7x-14}{1}}\\ & =15-5x+7x-14\\ & =2x+1\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Polinom}f(x)\text{dibagi oleh}(2x-4)\text{bersisa 6},\\ & \text{dibagi oleh}(x+4)\text{bersisa 24}.\\ & \text{Dan polinom}g(x)\text{dibagi oleh}(2x-4)\text{bersisa 5},\\ & \text{dibagi oleh}(x+4)\text{bersisa 2}.\\ & \text{Jika}h(x)=f(x).g(x),\text{maka}h(x)\\ & \text{dibagi}(2x^2+4x-16)\text{akan sisa}....\\ & \begin{array}{l}\\ \text{a}.-3x+24& & \text{d}.-6x+36\\ \text{b}.-3x+36& \text{c}.6x+24& \text{e}.12x+3\end{array}\\ \\ & \text{Jawab}:\\ & \text{Langkah pertama}\\ & \begin{array}{rl}f(x)& =(2x-4).h(x{)}_1+6\\ f(x)& =(x+4).h(x{)}_2+24\\ f(x)& =(2x-4)(x+4).H_1(x)+p_{1}x+q_1\\ & \text{Gunakanlah cara sebagai mana}\\ & \text{contoh soal No. 12 di atas yang}\\ \text{Alte}& \text{natif 2}\\ \text{mak}& \text{a}{p}_{1}x+q_1=-3x+12\end{array}\\ & \text{Langkah kedua}\\ & \begin{array}{rl}g(x)& =(2x-4).h(x{)}_3+5\\ g(x)& =(x+4).h(x{)}_4+2\\ g(x)& =(2x-4)(x+4).H_2(x)+p_{2}x+q_2\\ & \text{Gunakanlah cara sebagai mana}\\ & \text{contoh soal No. 12 di atas yang}\\ \text{Alte}& \text{natif 2}\\ \text{mak}& \text{a}{p}_{2}x+q_2={\displaystyle \frac{1}{2}x+4}\end{array}\\ & \text{Langkah ketiga}\\ & \begin{array}{rl}& h(x)=f(x)\times g(x)\\ & =((2x-4)(x+4)H_1(x)+(-3x+12))\\ & \times ((2x-4)(x+4)H_2(x)+{\displaystyle \frac{1}{2}x+4})\\ & \text{maka}\\ & \bullet h(2)=(0+(-3.2+12))(0+{\displaystyle \frac{1}{2}.2+4})=6.5=30\\ & \bullet h(-4)=(0+(-3.-4+12))(0+{\displaystyle \frac{1}{2}.-4+4})=24.2=48\\ & \text{Dengan pembagi}2x^2+x-16,\text{maka sisanya}:s_3(x)=p_{3}x+q_3\\ & \text{saat}x=2\Rightarrow 2p+q=30\\ & \text{saat}x=-4\Rightarrow -4p+q=48\\ & \text{selanjutnya dengan eliminasi-substitusi diperoleh}p=-3,q=36\\ & \text{sehingga}s(x)=px+q=-3x+36\end{array}\end{array}$