Contoh Soal 2 Polinom

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

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

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

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

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

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: (m-2)\: \: \textrm{adalah faktor dari}\: \: 2m^{3}+3tm+4,\\ &\textrm{maka nilai}\: \: t\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{10}{3}&&\textrm{d}.\quad -\displaystyle \frac{3}{10}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{3}&\textrm{c}.\quad \displaystyle \frac{3}{10}&\textrm{e}.\quad \color{red}-\displaystyle \frac{10}{3} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(m)&=2m^{3}+3tm+4\\ f(2)&=2(2)^{3}+3t(2)+4\\ 0&=16+6t+4\\ -6t&=20\\ t&=\color{red}-\displaystyle \frac{10}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{(KSM MA Kab/Kota 2015)Nilai terkecil}\: \: n\\ & \textrm{yang mengkin sehingga}\: \: n.(n+1).(n+2)\\\ & \textrm{habis dibagi 24 adalah}....\\ &\begin{array}{l}\\ \textrm{a}.\quad 1\\ \textrm{b}.\quad \color{red}2\\ \textrm{c}.\quad 3\\ \textrm{d}.\quad 4 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}k&=\displaystyle \frac{n.(n+1).(n+2)}{24}\\ &=\displaystyle \frac{n.(n+1).(n+2)}{2.(2+1).(2+2)}\\ &\textrm{maka}\: \: n=\color{red}2 \end{aligned} \end{array}$