$\begin{array}{ll}\\ 26.&\textrm{Banyaknya akar rasional bulat}\\ &\textrm{dari}\: \: 4x^{4}-15x^{2}+5x+6=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0&&\textrm{d}.\quad 3\\ \textrm{b}.\quad 1&\qquad&\textrm{e}.\quad 4\\ \textrm{c}.\quad \color{red}2\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{4x^{4}-15x^{2}+5x+6}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}4x^{3}+4x^{2}-11x-6&\color{blue}\textbf{hasil}\\ \hline \quad x-1&4x^{4}-15x^{2}+5x+6&\\ &4x^{4}-4x^{3}&-\\\hline &\: \: \: 4x^{3}-15x^{2}+5x+6&\\ &\: \: \: 4x^{3}-4x^{2}&- \\\hline &\: \: \quad\quad -11x^{2}+5x+6\\ &\: \: \quad\quad -11x^{2}+11x&-\\\hline &\quad \quad \qquad\qquad -6x+6\\ &\quad \quad \qquad\qquad -6x+6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=4x^{4}-15x^{2}+5x+6\\ &=\color{red}(x-1)\color{black}(4x^{3}+4x^{2}-11x-6) \end{aligned}\\\hline \end{array}\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Selanjutnya perhatikan pula }\\ &\displaystyle \frac{4x^{3}+4x^{2}-11x-6}{(x+2)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}4x^{2}-4x-3&\color{blue}\textbf{hasil}\\ \hline \quad x+2&4x^{3}+4x^{2}-11x-6&\\ &4x^{3}+8x^{2}&-\\\hline &\: \: \quad -4x^{2}-11x-6&\\ &\: \: \quad -4x^{2}-8x&- \\\hline &\: \: \qquad\qquad -3x-6\\ &\: \: \qquad\qquad -3x-6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad\quad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=4x^{3}+4x^{2}-11x-6\\ &=(x+2)(4x^{2}-4x-3)\\ &=\color{red}(x+2)\color{black}(2x-3)(2x+1) \end{aligned}\\\hline \end{array} \end{array}$
$\begin{array}{ll}\\ 27.&\textrm{Salah satu akar dari polinomial}\\ & 2x^{3}+5x^{2}+x-2=0\: \: \textrm{adalah}\: \: \displaystyle \frac{1}{2}\\ &\textrm{Jumlah dua akar yang lain adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6&&\textrm{d}.\quad \color{red}-3\\ \textrm{b}.\quad 3&\qquad&\textrm{e}.\quad -6\\ \textrm{c}.\quad 2\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{2x^{3}+5x^{2}+x-2}{(2x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}+3x+2&\color{blue}\textbf{hasil}\\ \hline \quad 2x-1&2x^{3}+5x^{2}+x-2&\\ &2x^{3}-x^{2}&-\\\hline &\: \: \: \qquad 6x^{2}+x-2&\\ &\: \: \: \qquad 6x^{2}-3x&- \\\hline &\: \: \: \: \: \qquad\qquad 4x-2\\ &\: \: \: \: \: \qquad\qquad 4x-2&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad\quad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=2x^{3}+5x^{2}+x-2\\ &=(2x-1)\color{red}(x^{2}+3x+2)\\ &=(2x-1)\color{red}(x+1)(x+2) \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga}\: \: x_{2}=-1,\: x_{3}=-2,\: \textrm{maka}\\ &x_{2}+x_{3}=-1+(-2)=-3\\ &\color{red}\textrm{atau dapat juga ditentukan dengan rumus}\\ &\textrm{ABC pada persamaan kuadrat, yaitu}\\ &x^{2}+3x+2=a_{2}x^{2}+a_{1}x+a_{0}\\ &\Leftrightarrow x^{2}+3x+2=-\displaystyle \frac{a_{1}}{a_{1}}=-\displaystyle \frac{3}{1}=-3 \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Salah satu akar dari polinomial}\\ & x^{4}-5x^{3}+5x^{2}+5x-6=0\\ &\textrm{adalah}\: \: 2\: .\: \textrm{Jumlah akar-akar yang }\\ &\textrm{lain adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 6&&\textrm{d}.\quad \color{red}3\\ \textrm{b}.\quad 5&\qquad&\textrm{e}.\quad 2\\ \textrm{c}.\quad 4\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{4}-5x^{3}+5x^{2}+5x-6}{(x-2)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{3}-3x^{2}-x+3&\color{blue}\textbf{hasil}\\ \hline \quad x-2&x^{4}-5x^{3}+5x^{2}+5x-6&\\ &x^{4}-2x^{3}&-\\\hline &\: \: \quad -3x^{3}+5x^{2}+5x-6&\\ &\: \: \quad -3x^{3}+6x^{2}&- \\\hline &\: \: \qquad\qquad -x^{2}+5x-6\\ &\: \: \qquad\qquad -x^{2}+2x&-\\\hline &\: \: \quad \qquad\qquad\qquad 3x-6\\ &\: \: \quad \qquad\qquad\qquad 3x-6&-\\\hline \qquad\textbf{Sisa}&\: \: \: \: \qquad\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=x^{4}-5x^{3}+5x^{2}+5x-6\\ &=(x-2)\color{red}(x^{3}-3x^{2}-x+3) \end{aligned}\\\hline \end{array} \\ &\textrm{Sehingga}\\ &x^{3}-3x^{2}-x+3=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ &\Leftrightarrow x_{1}+x_{2}+x_{3}=-\displaystyle \frac{a_{2}}{a_{2}}=-\frac{-3}{1}=3 \end{array}$.
$\begin{array}{ll}\\ 29.&\textrm{Akar-akar dari persamaan polinom}\\ & x^{3}-3x^{2}+4x+5=0\: \: \textrm{adalah}\: \: x_{1},x_{2}\\ &\textrm{dan}\: \: x_{3}.\: \: \textrm{Nilai}\: \: x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}1&&\textrm{d}.\quad 17\\ \textrm{b}.\quad 3&\qquad&\textrm{e}.\quad 19\\ \textrm{c}.\quad 9\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&x^{3}-3x^{2}+4x+5=ax^{3}+bx^{2}+cx+d\\ &\textrm{Perhatikan bahwa}:\\ &x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\\ &=(x_{1}+x_{2}+x_{3})^{2}-2(x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3})\\ &=\left (\displaystyle \frac{-b}{a} \right )^{2}-2\left ( \displaystyle \frac{c}{a} \right )\\ &=\left ( \displaystyle \frac{-(-3)}{1} \right )^{2}-2\left ( \displaystyle \frac{4}{1} \right )=9-8=1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textrm{Diketahui persamaan polinom}\\ & x^{3}-4x^{2}+6x-12=0\: \: \textrm{mempunyai}\\ &\textrm{akar-akar}\: \: x_{1},x_{2}\: \: \textrm{dan}\: \: x_{3}.\\ &\textrm{Nilai}\: \: \displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -3&&\textrm{d}.\quad \displaystyle \color{red}\frac{1}{2}\\ \textrm{b}.\quad -2&\qquad&\textrm{e}.\quad 3\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&x^{3}-4x^{2}+6x-12=ax^{3}+bx^{2}+cx+d\\ &\textrm{Perhatikan bahwa}:\\ &\displaystyle \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\\ &=\displaystyle \frac{x_{2}x_{3}+x_{1}x_{3}+x_{1}x_{2}}{x_{1}x_{2}x_{3}}\\ &=\displaystyle \frac{\displaystyle \frac{c}{a}}{-\displaystyle \frac{d}{a}}=-\frac{c}{d}=-\displaystyle \frac{6}{-12}=\frac{1}{2} \end{aligned} \end{array}$.
Tidak ada komentar:
Posting Komentar
Informasi