Contoh Soal Polinom (Bagian 5)

$\begin{array}{l}\\ 21.& \text{Akar yang mungkin dari persamaan}\\ & 4x^3-p{x}^2+qx-6=0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{6}}& & \text{d}.{\displaystyle \frac{3}{2}}\\ \text{b}.{\displaystyle \frac{2}{3}}& & \text{e}.4\\ \text{c}.{\displaystyle \frac{4}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui}4x^3-p{x}^2+qx-6=0\\ \text{Deng}& \text{an teorema faktor, faktor yang}\\ \text{mung}& \text{kin adalah}{\displaystyle \frac{6}{4}=\pm 1,\pm {\displaystyle \frac{3}{2}}}\end{array}\\ & \end{array}$.

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

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

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

$\begin{array}{l}\\ 25.& \text{Banyaknya akar rasional dari}\\ & x^4-3x^2+2=0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.4& & \text{d}.1\\ \text{b}.3& & \text{e}.0\\ \text{c}.2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui}{x}^4-3x^2+2=0\\ \iff & {\left(x^2\right)}^2-3\left(x^2\right)+2=0\\ \iff & (x^2-1)(x^2-2)=0\end{array}\\ & \text{Maka akar-akarnya}\\ & x^4-3x^2+2\\ & =(x^2-1)(x^2-2)\\ & =(x+1)(x-1)(x-\sqrt{2})(x+\sqrt{2})\end{array}$.