$\begin{array}{ll}\\ 21.&\textrm{Akar yang mungkin dari persamaan}\\ &4x^{3}-px^{2}+qx-6=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{6}&&\textrm{d}.\quad \color{red}\displaystyle \frac{3}{2}\\ \textrm{b}.\quad \displaystyle \frac{2}{3}&\qquad&\textrm{e}.\quad 4\\ \textrm{c}.\quad \displaystyle \frac{4}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: \color{red}4x^{3}-px^{2}+qx-6=0\\ \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: \displaystyle \frac{6}{4}=\pm 1,\pm \displaystyle \frac{3}{2} \end{aligned}\\ & \end{array}$.
$\begin{array}{ll}\\ 22.&\textrm{Akar-akar dari suku banyak berikut}\\ &f(x)=x^{3}-2x^{2}-5x+6\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-2,1\: \: \textrm{dan}\: \: 3&&\\ \textrm{b}.\quad -2,-1\: \: \textrm{dan}\: \: 3&&\\ \textrm{c}.\quad -3,1\: \: \textrm{dan}\: \: 3\\ \textrm{d}.\quad -3,-1\: \: \textrm{dan}\: \: 2\\ \textrm{e}.\quad \color{red}-2,1\: \: \textrm{dan}\: \: 3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: f(x)=\color{blue}x^{3}-2x^{2}-5x+6\\ \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: 6=\pm 1,\pm 2,\pm 3,\pm 6\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(1)&=(1)^{3}-2(1)^{2}-5(1)+6=0\\ \textrm{maka}&\: \: \color{red}x-1\: \: \color{black}\textrm{termasuk faktornya} \end{aligned}\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{3}-2x^{2}-5x+6}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}-x-6\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-1&x^{3}-2x^{2}-5x+6&\\ &x^{3}-x^{2}&-\\\hline &\: \: \: \quad -x^{2}-5x+6&\\ &\: \: \: \quad -x^{2}+x&- \\\hline &\: \: \qquad\quad\quad -6x+6\\ &\: \: \qquad\quad\quad -6x+6&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=x^{3}-2x^{2}-5x+6\\ &=(x-1)(x^{2}-x-6)\\ &=\color{red}(x-3)\color{black}(x-1)\color{red}(x+2) \end{aligned}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 23.&\textrm{Akar-akar rasional suku banyak dari}\\ &2x^{3}+5x^{2}-4x-3=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1,\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: 3&&\\ \textrm{b}.\quad 1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: 3&&\\ \textrm{c}.\quad \color{red}1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: -3\\ \textrm{d}.\quad -1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: -3\\ \textrm{e}.\quad -1,-\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \: 3 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: f(x)=\color{blue}2x^{3}+5x^{2}-4x-3\\ \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: \displaystyle \frac{3}{2}=\pm 1,\pm \frac{3}{2}\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(1)&=2.(1)^{3}+5.(1)^{2}-4(1)-3=0\\ \textrm{maka}&\: \: \color{red}x-1\: \: \color{black}\textrm{termasuk faktornya} \end{aligned}\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{2x^{3}+5x^{2}-4x-3}{(x-1)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}2x^{2}+7x+3\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-1&2x^{3}+5x^{2}-4x-3&\\ &2x^{3}-2x^{2}&-\\\hline &\: \: \: \quad 7x^{2}-4x-3&\\ &\: \: \: \quad 7x^{2}-7x&- \\\hline &\: \: \qquad\quad\quad 3x-3\\ &\: \: \qquad\quad\quad 3x-3&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \qquad f(x)&=2x^{3}+5x^{2}-4x-3\\ &=(x-1)(2x^{2}+7x+3)\\ &=\color{red}(2x+1)\color{black}(x-1)\color{red}(x+3) \end{aligned}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 24.&\textrm{Akar-akar rasional suku banyak dari}\\ &f(x)=x^{3}-6x^{2}+9x-2\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2,1-\sqrt{3}\: \: \textrm{dan}\: \: 1+\sqrt{3}&&\\ \textrm{b}.\quad 1,2-\sqrt{3}\: \: \textrm{dan}\: \: 2+\sqrt{3}&&\\ \textrm{c}.\quad -1,2-\sqrt{3}\: \: \textrm{dan}\: \: 3+\sqrt{3}\\ \textrm{d}.\quad \color{red}2,2-\sqrt{3}\: \: \textrm{dan}\: \: 2+\sqrt{3}\\ \textrm{e}.\quad 1,1-\sqrt{3}\: \: \textrm{dan}\: \: 1+\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: f(x)=\color{blue}x^{3}-6x^{2}+9x-2\\ \textrm{Deng}&\textrm{an teorema faktor, faktor yang}\\ \textrm{mung}&\textrm{kin adalah}\: \: 2=\pm 1,\pm 2\\ \textrm{Deng}&\textrm{an substitusi akan diperoleh}\\ f(2)&=(2)^{3}-6.(2)^{2}+9(2)-2=0\\ \textrm{maka}&\: \: \color{red}x-1\: \: \color{black}\textrm{termasuk faktornya} \end{aligned}\\ &\begin{array}{|l|}\hline \begin{aligned} &\textrm{Perhatikan uraian berikut}\\ &\displaystyle \frac{x^{3}-6x^{2}+9x-2}{(x-2)}\\ \end{aligned}\\\\ \begin{array}{l|lll}\: \: \: \textbf{pembagi}&\quad \color{red}x^{2}-4x+1\quad \color{blue}\textbf{hasil}&\color{blue}\textbf{bagi}\\ \hline \quad x-2&x^{3}-6x^{2}+9x-2&\\ &x^{3}-2x^{2}&-\\\hline &\: \: \: \quad -4x^{2}+9x-2&\\ &\: \: \: \quad -4x^{2}+8x&- \\\hline &\: \: \qquad\qquad\quad x-2\\ &\: \: \qquad\qquad\quad x-2&-\\\hline \qquad\textbf{Sisa}&\qquad\qquad\qquad 0& (\textbf{habis}) \end{array}\\\\ \begin{aligned}\therefore \quad f(x)&=x^{3}-6x^{2}+9x-2\\ &=(x-2)(x^{2}-4x+1)\\ &=\color{red}(x-2-\sqrt{3})\color{black}(x-2)\color{red}(x-2+\sqrt{3}) \end{aligned}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 25.&\textrm{Banyaknya akar rasional dari}\\ &x^{4}-3x^{2}+2=0\: \: \textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}4&&\textrm{d}.\quad 1\\ \textrm{b}.\quad 3&\qquad&\textrm{e}.\quad 0\\ \textrm{c}.\quad 2\\ \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\: \: \color{blue}x^{4}-3x^{2}+2=0\\ \Leftrightarrow &\: \: \left (x^{2} \right )^{2}-3\left ( x^{2} \right )+2=0\\ \Leftrightarrow &\: \: \left ( x^{2}-1 \right )\left ( x^{2}-2 \right )=0 \end{aligned}\\ &\textrm{Maka akar-akarnya}\\ &x^{4}-3x^{2}+2\\ &=\left ( x^{2}-1 \right )\left ( x^{2}-2 \right )\\ &=\color{red}(x+1)(x-1)(x-\sqrt{2})(x+\sqrt{2}) \end{array}$.
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