Contoh Soal Polinom (Bagian 6)

$\begin{array}{ll}\\ 26.&\textrm{Banyaknya akar rasional bulat}\\ &\textrm{dari}\: \: 4x^{4}-15x^{2}+5x+6=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0&&\textrm{d}.\quad 3\\ \textrm{b}.\quad  1&\qquad&\textrm{e}.\quad 4\\ \textrm{c}.\quad  \color{red}2\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{4x^{4}-15x^{2}+5x+6}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}4x^{3}+4x^{2}-11x-6&\color{blue}\textbf{hasil}\\ \hline \quad x-1&4x^{4}-15x^{2}+5x+6&\\ &4x^{4}-4x^{3}&-\\\hline &\: \: \:    4x^{3}-15x^{2}+5x+6&\\ &\: \: \: 4x^{3}-4x^{2}&- \\\hline &\: \: \quad\quad -11x^{2}+5x+6\\ &\: \: \quad\quad -11x^{2}+11x&-\\\hline &\quad \quad \qquad\qquad -6x+6\\ &\quad \quad \qquad\qquad -6x+6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=4x^{4}-15x^{2}+5x+6\\ &=\color{red}(x-1)\color{black}(4x^{3}+4x^{2}-11x-6) \end{aligned}\\\hline \end{array}\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Selanjutnya perhatikan pula }\\ &\displaystyle \frac{4x^{3}+4x^{2}-11x-6}{(x+2)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}4x^{2}-4x-3&\color{blue}\textbf{hasil}\\ \hline \quad x+2&4x^{3}+4x^{2}-11x-6&\\ &4x^{3}+8x^{2}&-\\\hline &\: \: \quad    -4x^{2}-11x-6&\\ &\: \: \quad -4x^{2}-8x&- \\\hline &\: \: \qquad\qquad -3x-6\\ &\: \: \qquad\qquad -3x-6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad\quad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=4x^{3}+4x^{2}-11x-6\\ &=(x+2)(4x^{2}-4x-3)\\ &=\color{red}(x+2)\color{black}(2x-3)(2x+1) \end{aligned}\\\hline \end{array}   \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Salah satu akar dari polinomial}\\ & 2x^{3}+5x^{2}+x-2=0\: \: \textrm{adalah}\: \: \displaystyle \frac{1}{2}\\ &\textrm{Jumlah dua akar yang lain adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6&&\textrm{d}.\quad \color{red}-3\\ \textrm{b}.\quad  3&\qquad&\textrm{e}.\quad -6\\ \textrm{c}.\quad  2\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{2x^{3}+5x^{2}+x-2}{(2x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}+3x+2&\color{blue}\textbf{hasil}\\ \hline \quad 2x-1&2x^{3}+5x^{2}+x-2&\\ &2x^{3}-x^{2}&-\\\hline &\: \: \: \qquad   6x^{2}+x-2&\\ &\: \: \: \qquad 6x^{2}-3x&- \\\hline &\: \: \: \: \: \qquad\qquad 4x-2\\ &\: \: \: \: \: \qquad\qquad 4x-2&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad\quad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=2x^{3}+5x^{2}+x-2\\ &=(2x-1)\color{red}(x^{2}+3x+2)\\ &=(2x-1)\color{red}(x+1)(x+2) \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga}\: \: x_{2}=-1,\: x_{3}=-2,\: \textrm{maka}\\ &x_{2}+x_{3}=-1+(-2)=-3\\ &\color{red}\textrm{atau dapat juga ditentukan dengan rumus}\\ &\textrm{ABC pada persamaan kuadrat, yaitu}\\ &x^{2}+3x+2=a_{2}x^{2}+a_{1}x+a_{0}\\ &\Leftrightarrow x^{2}+3x+2=-\displaystyle \frac{a_{1}}{a_{1}}=-\displaystyle \frac{3}{1}=-3 \end{array}$.

$\begin{array}{ll}\\ 28.&\textrm{Salah satu akar dari polinomial}\\ & x^{4}-5x^{3}+5x^{2}+5x-6=0\\ &\textrm{adalah}\: \: 2\: .\: \textrm{Jumlah akar-akar yang }\\ &\textrm{lain adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6&&\textrm{d}.\quad \color{red}3\\ \textrm{b}.\quad  5&\qquad&\textrm{e}.\quad 2\\ \textrm{c}.\quad  4\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{4}-5x^{3}+5x^{2}+5x-6}{(x-2)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{3}-3x^{2}-x+3&\color{blue}\textbf{hasil}\\ \hline \quad x-2&x^{4}-5x^{3}+5x^{2}+5x-6&\\ &x^{4}-2x^{3}&-\\\hline &\: \:  \quad   -3x^{3}+5x^{2}+5x-6&\\ &\: \:  \quad -3x^{3}+6x^{2}&- \\\hline &\: \:    \qquad\qquad -x^{2}+5x-6\\ &\: \:   \qquad\qquad -x^{2}+2x&-\\\hline &\: \: \quad    \qquad\qquad\qquad 3x-6\\ &\: \: \quad  \qquad\qquad\qquad 3x-6&-\\\hline \qquad\textbf{Sisa}&\: \: \: \: \qquad\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=x^{4}-5x^{3}+5x^{2}+5x-6\\ &=(x-2)\color{red}(x^{3}-3x^{2}-x+3) \end{aligned}\\\hline \end{array} \\ &\textrm{Sehingga}\\ &x^{3}-3x^{2}-x+3=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ &\Leftrightarrow x_{1}+x_{2}+x_{3}=-\displaystyle \frac{a_{2}}{a_{2}}=-\frac{-3}{1}=3 \end{array}$.

$\begin{array}{ll}\\ 29.&\textrm{Akar-akar dari persamaan polinom}\\ & x^{3}-3x^{2}+4x+5=0\: \: \textrm{adalah}\: \: x_{1},x_{2}\\ &\textrm{dan}\: \: x_{3}.\: \: \textrm{Nilai}\: \: x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}1&&\textrm{d}.\quad 17\\ \textrm{b}.\quad  3&\qquad&\textrm{e}.\quad 19\\ \textrm{c}.\quad  9\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&x^{3}-3x^{2}+4x+5=ax^{3}+bx^{2}+cx+d\\ &\textrm{Perhatikan bahwa}:\\ &x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\ &=(x_{1}+x_{2}+x_{3})^{2}-2(x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3})\\ &=\left (\displaystyle \frac{-b}{a}  \right )^{2}-2\left ( \displaystyle \frac{c}{a} \right )\\ &=\left ( \displaystyle \frac{-(-3)}{1} \right )^{2}-2\left ( \displaystyle \frac{4}{1} \right )=9-8=1 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Diketahui persamaan polinom}\\ & x^{3}-4x^{2}+6x-12=0\: \: \textrm{mempunyai}\\ &\textrm{akar-akar}\: \: x_{1},x_{2}\: \: \textrm{dan}\: \: x_{3}.\\ &\textrm{Nilai}\: \: \displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -3&&\textrm{d}.\quad \displaystyle \color{red}\frac{1}{2}\\ \textrm{b}.\quad  -2&\qquad&\textrm{e}.\quad 3\\ \textrm{c}.\quad  \displaystyle \frac{1}{3}\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&x^{3}-4x^{2}+6x-12=ax^{3}+bx^{2}+cx+d\\ &\textrm{Perhatikan bahwa}:\\  &\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\\ &=\displaystyle \frac{x_{2}x_{3}+x_{1}x_{3}+x_{1}x_{2}}{x_{1}x_{2}x_{3}}\\ &=\displaystyle \frac{\displaystyle \frac{c}{a}}{-\displaystyle \frac{d}{a}}=-\frac{c}{d}=-\displaystyle \frac{6}{-12}=\frac{1}{2} \end{aligned}  \end{array}$.

Contoh Soal Polinom (Bagian 5)

$\begin{array}{ll}\\ 21.&\textrm{Akar yang mungkin dari persamaan}\\ &4x^{3}-px^{2}+qx-6=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{6}&&\textrm{d}.\quad \color{red}\displaystyle \frac{3}{2}\\ \textrm{b}.\quad  \displaystyle \frac{2}{3}&\qquad&\textrm{e}.\quad 4\\ \textrm{c}.\quad  \displaystyle \frac{4}{3}\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: \color{red}4x^{3}-px^{2}+qx-6=0\\  \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: \displaystyle \frac{6}{4}=\pm 1,\pm \displaystyle \frac{3}{2} \end{aligned}\\ & \end{array}$.

$\begin{array}{ll}\\ 22.&\textrm{Akar-akar dari suku banyak berikut}\\ &f(x)=x^{3}-2x^{2}-5x+6\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-2,1\: \: \textrm{dan}\: \: 3&&\\ \textrm{b}.\quad  -2,-1\: \: \textrm{dan}\: \: 3&&\\ \textrm{c}.\quad  -3,1\: \: \textrm{dan}\: \: 3\\ \textrm{d}.\quad  -3,-1\: \: \textrm{dan}\: \: 2\\ \textrm{e}.\quad  \color{red}-2,1\: \: \textrm{dan}\: \: 3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: f(x)=\color{blue}x^{3}-2x^{2}-5x+6\\  \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: 6=\pm 1,\pm 2,\pm 3,\pm 6\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(1)&=(1)^{3}-2(1)^{2}-5(1)+6=0\\ \textrm{maka}&\: \: \color{red}x-1\: \: \color{black}\textrm{termasuk faktornya} \end{aligned}\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{3}-2x^{2}-5x+6}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}-x-6\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-1&x^{3}-2x^{2}-5x+6&\\ &x^{3}-x^{2}&-\\\hline &\: \: \:   \quad -x^{2}-5x+6&\\ &\: \: \:  \quad -x^{2}+x&- \\\hline &\: \: \qquad\quad\quad -6x+6\\ &\: \: \qquad\quad\quad -6x+6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=x^{3}-2x^{2}-5x+6\\ &=(x-1)(x^{2}-x-6)\\ &=\color{red}(x-3)\color{black}(x-1)\color{red}(x+2) \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Akar-akar rasional suku banyak dari}\\ &2x^{3}+5x^{2}-4x-3=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1,\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: 3&&\\ \textrm{b}.\quad  1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: 3&&\\ \textrm{c}.\quad  \color{red}1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: -3\\ \textrm{d}.\quad  -1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: -3\\ \textrm{e}.\quad  -1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: 3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: f(x)=\color{blue}2x^{3}+5x^{2}-4x-3\\  \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: \displaystyle \frac{3}{2}=\pm 1,\pm \frac{3}{2}\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(1)&=2.(1)^{3}+5.(1)^{2}-4(1)-3=0\\ \textrm{maka}&\: \: \color{red}x-1\: \: \color{black}\textrm{termasuk faktornya} \end{aligned}\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{2x^{3}+5x^{2}-4x-3}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}2x^{2}+7x+3\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-1&2x^{3}+5x^{2}-4x-3&\\ &2x^{3}-2x^{2}&-\\\hline &\: \: \:   \quad 7x^{2}-4x-3&\\ &\: \: \:  \quad 7x^{2}-7x&- \\\hline &\: \: \qquad\quad\quad 3x-3\\ &\: \: \qquad\quad\quad 3x-3&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=2x^{3}+5x^{2}-4x-3\\ &=(x-1)(2x^{2}+7x+3)\\ &=\color{red}(2x+1)\color{black}(x-1)\color{red}(x+3) \end{aligned}\\\hline \end{array}  \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Akar-akar rasional suku banyak dari}\\ &f(x)=x^{3}-6x^{2}+9x-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2,1-\sqrt{3}\: \: \textrm{dan}\: \: 1+\sqrt{3}&&\\ \textrm{b}.\quad  1,2-\sqrt{3}\: \: \textrm{dan}\: \: 2+\sqrt{3}&&\\ \textrm{c}.\quad  -1,2-\sqrt{3}\: \: \textrm{dan}\: \: 3+\sqrt{3}\\ \textrm{d}.\quad  \color{red}2,2-\sqrt{3}\: \: \textrm{dan}\: \: 2+\sqrt{3}\\ \textrm{e}.\quad  1,1-\sqrt{3}\: \: \textrm{dan}\: \: 1+\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: f(x)=\color{blue}x^{3}-6x^{2}+9x-2\\  \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: 2=\pm 1,\pm 2\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(2)&=(2)^{3}-6.(2)^{2}+9(2)-2=0\\ \textrm{maka}&\: \: \color{red}x-1\: \: \color{black}\textrm{termasuk faktornya} \end{aligned}\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{3}-6x^{2}+9x-2}{(x-2)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}-4x+1\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-2&x^{3}-6x^{2}+9x-2&\\ &x^{3}-2x^{2}&-\\\hline &\: \: \:   \quad -4x^{2}+9x-2&\\ &\: \: \:  \quad -4x^{2}+8x&- \\\hline &\: \: \qquad\qquad\quad x-2\\ &\: \: \qquad\qquad\quad x-2&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=x^{3}-6x^{2}+9x-2\\ &=(x-2)(x^{2}-4x+1)\\ &=\color{red}(x-2-\sqrt{3})\color{black}(x-2)\color{red}(x-2+\sqrt{3}) \end{aligned}\\\hline \end{array}  \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Banyaknya akar rasional dari}\\ &x^{4}-3x^{2}+2=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}4&&\textrm{d}.\quad 1\\ \textrm{b}.\quad  3&\qquad&\textrm{e}.\quad 0\\ \textrm{c}.\quad  2\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: \color{blue}x^{4}-3x^{2}+2=0\\  \Leftrightarrow &\: \: \left (x^{2}  \right )^{2}-3\left ( x^{2} \right )+2=0\\ \Leftrightarrow &\: \: \left ( x^{2}-1 \right )\left ( x^{2}-2 \right )=0 \end{aligned}\\ &\textrm{Maka akar-akarnya}\\ &x^{4}-3x^{2}+2\\ &=\left ( x^{2}-1 \right )\left ( x^{2}-2 \right )\\ &=\color{red}(x+1)(x-1)(x-\sqrt{2})(x+\sqrt{2})   \end{array}$.

Ketaksamaan Minkowski

SELAMAT HARI RAYA IDUL FITRI 1443 H

 

Contoh Soal Polinom (Bagian 4)

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

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

$\begin{array}{ll}\\ 18.&\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}$.

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

$\begin{array}{ll}\\ 20.&\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}$

Contoh Soal Polinom (Bagian 3)

$\begin{array}{ll}\\ 11.&\textrm{Jika polinom}\: \: 2x^{3}+7x^{2}+ax-3\\  &\textrm{mempunyai faktor}\: \: 2x-1,\: \textrm{maka}\\ &\textrm{faktor linear lainnya adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (x-3)\: \: \textrm{dan}\: \: (x+1)&&\\ \textrm{b}.\quad  \color{red}(x+3)\: \: \textrm{dan}\: \: (x+1)&&\\ \textrm{c}.\quad  (x+3)\: \: \textrm{dan}\: \: (x-1)\\ \textrm{d}.\quad  (x-3)\: \: \textrm{dan}\: \: (x-1)\\ \textrm{e}.\quad  (x+2)\: \: \textrm{dan}\: \: (x-6) \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{2x^{3}+7x^{2}+2x-3}{(2x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}+4x+3\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad 2x-1&2x^{3}+7x^{2}+2x-3&\\ &2x^{3}-x^{2}&-\\\hline &\: \: \:   \qquad 8x^{2}+2x-3&\\ &\: \: \:  \qquad 8x^{2}-4x&- \\\hline &\: \: \qquad\qquad\quad 6x-3\\ &\: \: \qquad\qquad\quad 6x-3&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=2x^{3}+7x^{2}+2x-3\\ &=(2x-1)(x^{2}+4x+3)\\ &=(2x-1)\color{red}(x+1)(x+3) \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Diketahui}\: \: g(x)=2x^{3}+ax^{2}+bx+6\\  &h(x)=x^{2}+x-6\: \: \textrm{adalah faktor dari}\\ &g(x)\: ,\: \textrm{Nilai}\: \: a\: \: \textrm{yang memenuhi adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -3&&\textrm{d}.\quad 2\\ \textrm{b}.\quad  -1&\textrm{c}.\quad  1&\textrm{e}.\quad \color{red}5 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \: g(x)=2x^{3}+ax^{2}+bx+6\\ &\textrm{dengan pembagi}\: \: h(x)=x^{2}+x-6\\ &\Leftrightarrow \: \: h(x)=(x+3)(x-2)\\ &\textrm{Hal ini artinya}\\ &g(-3)=2(-3)^{3}+a(-3)^{2}+b(-3)+6\\ &\: \: \: \qquad =-54+9a-3b+6=0\: ....(1)\\ &g(2)=2(2)^{3}+a(2)^{2}+b(2)+6\\ &\: \: \: \: \quad =16+4a+2b+6=0\: ..........(2)\\ &\textrm{Dengan mengeliminasi persamaan}\\ &(1)\: \: \textrm{dengan persamaan}\: \: (2),\: \textrm{maka}\\ & \end{aligned}\\ &\begin{array}{llllll} g(-3)&=&9a-3b&=&48\\ g(2)&=&4a+2b&=&-22&\\\hline (x2)&&18a-6b&=&96\\ (x3)&&12a+6b&=&-66&+\\\hline &&6a&=&30\\ &&\quad\qquad a&=&5 \end{array} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: f(x)=(x-1)(x+1)(x-2)\\ &\textrm{maka berikut yang bukan faktor}\\ &f(-x)\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (x-1)&&\textrm{d}.\quad (x+2)\\ \textrm{b}.\quad  (x+1)&\textrm{c}.\quad  \color{red}(x-2)&\textrm{e}.\quad (1-x) \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \: f(x)=(x-1)(x+1)(x-2)\\ &\Leftrightarrow f(-x)=(-x-1)(-x+1)(-x-2)\\ &\Leftrightarrow f(-x)=(x+1)(-x+1)(x+2)\\ &\textrm{atau}\\ &\Leftrightarrow f(-x)=(-x-1)(x-1)(x+2)\\ &\textrm{atau}\\ &\Leftrightarrow f(-x)=(x+1)(x-1)(-x-2)\\ &\textrm{Perhatikan bahwa faktor}\\ &(x-2)\: \: \textrm{tidak akan pernah ada} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: n\: \: \textrm{merupakan bilangan bulat }\\ &\textrm{positif, pernyataan berikut ini}\\ &\textrm{yang benar adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{n}+1\: \: \textrm{habis dibagi}\: \: (x+1)&&\\ \textrm{b}.\quad  x^{n}+1\: \: \textrm{habis dibagi}\: \: (x-1)&&\\ \textrm{c}.\quad  x^{n}-1\: \: \textrm{habis dibagi}\: \: (x+1)\\ \textrm{d}.\quad  \color{red}x^{n}-1\: \: \textrm{habis dibagi}\: \: (x-1)\\ \textrm{e}.\quad  x^{n}+1\: \: \textrm{habis dibagi}\: \: (x+2) \end{array}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\bullet \quad x^{n}+1=(x+1)(x^{n-1}+1)-x(x^{n-2}+1)\\ &\bullet \quad x^{n}-1=\color{red}(x-1)\color{black}(x^{n-1}+x^{n-2}+\cdots +x+1) \end{aligned}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned}&\begin{array}{|c|c|l|}\hline \textrm{Polinom}&\textrm{Pembagi}&\textrm{Hasil dengan}\: \: n\: \: \textrm{positif}\\\hline x^{n}+1&x+1&f(-1)=(-1)^{n}+1=....\\\hline x^{n}+1&x-1&f(1)=(1)^{n}+1=2\\\hline x^{n}-1&x+1&f(-1)=(-1)^{n}-1=-2\\\hline x^{n}-1&\color{red}x-1&\color{red}f(1)=(1)^{n}-1=0\\\hline x^{n}+1&x+2&f(-2)=(-2)^{n}+1\neq 0\\\hline \end{array}\\ &\textrm{Sebagai catatan bahwa saat}\: \: \: \displaystyle \frac{x^{n}+1}{x+1}=....\\ &\bullet \quad\textrm{ketika}\: \: n=\textrm{ganjil, maka}\: \displaystyle \frac{x^{n}+1}{x+1}=0,\: \textrm{tetapi}\\ &\bullet \quad \textrm{ketika}\: \: n=\textrm{genap, maka}\: \displaystyle \frac{x^{n}+1}{x+1}\neq 0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika salah satu akar dari polinom}\\ &\: \: x^{3}+4x^{2}+x-6=0\: \: \textrm{adalah}\: \: x=1,\\  &\textrm{maka akar-akar yang lain adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2\: \: \textrm{dan}\: \: 3&&\\ \textrm{b}.\quad  -3\: \: \textrm{dan}\: \: 2&&\\ \textrm{c}.\quad  -2\: \: \textrm{dan}\: \: 3\\ \textrm{d}.\quad  \color{red}-3\: \: \textrm{dan}\: \: -2\\ \textrm{e}.\quad  1\: \: \textrm{dan}\: \: \displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline  \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{3}+4x^{2}+x-6}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}+5x+6\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-1&x^{3}+4x^{2}+x-6&\\ &x^{3}-x^{2}&-\\\hline &\: \: \:   \qquad 5x^{2}+x-6&\\ &\: \: \:  \qquad 5x^{2}-5x&- \\\hline &\: \: \qquad\qquad\quad 6x-6\\ &\: \: \qquad\qquad\quad 6x-6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=x^{3}+4x^{2}+x-6\\ &=(x-1)(x^{2}+5x+6)\\ &=(x-1)\color{red}(x+2)(x+3) \end{aligned}\\\hline \end{array} \end{array}$

Contoh Soal Polinom (Bagian 2)

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

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

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

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

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

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

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

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

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