Contoh Soal Polinom (Bagian 9)

 $\begin{array}{l}\\ 41.& \mathbf{\text{Mat OSN-Kab 2019}}\\ & \text{Semua bilangan bulat}n\text{sehingga}\\ & n^4+16n^3+71n^2+56n\text{merupakan}\\ & \text{bilangan kadrat tidak nol adalah}....\\ \end{array}$

DAFTAR PUSTAKA

1. Kartini, Suprapto, Subandi, Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
2. Sembiring. S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.
3. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Pemintan Matematika dan Ilmu-Ilmu Alam. Bandung: Srikandi Empat Widya Utama.
4. Sukino. 2016. Matematika Jilid 2 untuk Siswa SMA/MA Kelas XI Kelompok Pemintan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh Soal Polinom (Bagian 8)

$\begin{array}{l}\\ 36.& (\mathbf{\text{Soal Seleksi OM Af-Sel 1983}})\\ & \text{Jika diketahui}{a}^3-a-1=0\text{maka}\\ & a^4+a^3-a^2-2a+1=....\\ & \begin{array}{l}\\ \text{a}.0& & \text{d}.1\\ \text{b}.a+1& & \text{e}.a^3+a+1\\ \text{c}.2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Lihat pembahasan no}.35\\ & \text{Cukup jelas }\end{array}\end{array}$ .

$\begin{array}{l}\\ 37.& \text{Tentukanlah suku banyak}f(x)\text{sedemikian}\\ & \text{sehingga}f(x)\text{terbagi oleh}{x}^2+1,\\ & \text{sedangkan}f(x)+1\text{terbagi oleh}{x}^3+x^2+1\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =(x^2+1).h_{1}x\\ f(x)+1& =(x^2+1).h_{1}x+1\\ \text{supaya}& f(x)+1\text{terbagi habis oleh}{x}^3+x^2+1,\\ & \text{maka akan ada bilangan bulat}k,(k\ne 0)\\ k& ={\displaystyle \frac{f(x)+1}{x^3+x^2+1}}\\ & ={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}}\\ k=1\Rightarrow & 1={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}\text{maka}{h}_{1}x=x}\\ \text{sehingga}& f(x)=x^3+x^2\\ \text{untuk ni}& \text{lai}k\text{yang lain, tak ditemukan}\end{array}\end{array}$.

$\begin{array}{l}\\ 38.& (\text{KSM 2015})\text{Diketahui}f(x)\text{adalah polinom}\\ & (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ & \text{dengan}{x}_1,x_2,x_3,x_4,\text{dan}{x}_5\text{adalah}\\ & \text{bilangan bulat berbeda}.\text{Jika}f(104)=2012,\\ & \text{maka nilai}{x}_1+x_2+x_3+x_4+x_5\text{sama dengan}....\\ & \text{a}.13\\ & \text{b}.14\\ & \text{c}.16\\ & \text{d}.17\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}:\\ f(x)& =(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ f(104)& =(104-x_1)(104-x_2)(104-x_3)(104-x_4)(104-x_5)=2012\\ & =2012=1\times 2\times 503\\ & =(-1)\times (1)\times (-2)\times (2)\times (503)\\ \text{maka}& \{\begin{array}{ll}(104-x_1)& =-2\Rightarrow x_1=106\\ (104-x_2)& =-1\Rightarrow x_2=105\\ (104-x_3)& =1\Rightarrow x_3=103\\ (104-x_4)& =2\Rightarrow x_4=102\\ (104-x_5)& =503\Rightarrow x_5=-399\end{array}\\ \text{sehin}& \text{gga},\\ & x_1+x_2+x_3+x_4+x_5=106+105+103+102+(-399)=17\end{array}\end{array}$.

$\begin{array}{l}\\ 39.& \text{Akar-akar persamaan}\\ & (x+1)(2x+1)(3x-1)(4x-1)+6x^4=0\\ & \text{adalah}{x}_1,x_2,x_3\text{dan}{x}_4.\text{Jika}\\ & x_1<x_2<x_3<x_4\text{dan}{x}_1+x_4=m\\ & \text{serta}{x}_2+x_3=n,\text{maka}mn=....\\ & \begin{array}{l}\\ \text{a}.-{\displaystyle \frac{7}{30}}& & \text{d}.{\displaystyle \frac{2}{15}}\\ \text{b}.{\displaystyle \frac{7}{30}}& & \text{e}.-{\displaystyle \frac{4}{15}}\\ \text{c}.{\displaystyle \frac{4}{15}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & (x+1)(2x+1)(3x-1)(4x-1)+6x^4=0\\ & \iff (2x^2+3x+1)(12x^2-7x+1)+6x^4=0\\ & \iff 24x^4+22x^3-7x^2-4x+1+6x^4=0\\ & \iff 30x^4+22x^3-7x^2-4x+1+6x^4=0\\ & \iff (6x^2+2x+2)(5x^2+2x-1)=0\\ & \iff (6x^2+2x+2)=0\text{V}(5x^2+2x-1)=0\\ & \{\begin{array}{ll}{x}_1& ={\displaystyle \frac{-1-\sqrt{6}}{5}}\\ x_2& ={\displaystyle \frac{-1-\sqrt{7}}{6}}\end{array},\{\begin{array}{ll}{x}_3& ={\displaystyle \frac{-1+\sqrt{7}}{6}}\\ x_4& ={\displaystyle \frac{-1+\sqrt{6}}{5}}\end{array}\\ & \text{Dengan}m=x_1+x_4=-{\displaystyle \frac{2}{5},n=x_2+x_3=-{\displaystyle \frac{1}{3}}}\\ & \text{maka}mn=(-{\displaystyle \frac{2}{5}})(-{\displaystyle \frac{1}{3}})={\displaystyle \frac{2}{15}}\end{array}\end{array}$ 

$\begin{array}{l}\\ 40.& \text{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ & \text{adalah}{x}_1,x_2,x_3,...,x_{2017}\\ & \text{Tentukan nilai dari}\\ & {\displaystyle \frac{1}{1-x_1}+\frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{x^{2018}-1}{x-1}}& =x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ \text{perlu dii}& \text{ngat bahwa kondisi ini mensyaratkan}x\ne 1,\\ & \text{sehingga}\\ \\ x^{2018}-1& =0\\ x^{2018}& =1\\ x& =\pm 1,\text{pilih}x=-1\\ \text{maka}& \text{nilai dari}\\ {\displaystyle \frac{1}{1-x_1}+}& \frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+{\displaystyle \frac{1}{2}}}}}\\ & ={\displaystyle \frac{2017}{2}}\end{array}\end{array}$.

Contoh Soal Polinom (Bagian 7)

$\begin{array}{l}\\ 31.& \text{Diketahui faktor-faktor polinom}\\ & x^3+p{x}^2-3x+q=0\text{adalah}\\ & (x+2)\text{dan}(x-3).\text{Jika akar-akar}\\ & \text{polinom tersebut adalah}{x}_1,x_2\text{dan}{x}_3,\\ & \text{maka nilai}{x}_1+x_2+x_3=....\\ & \begin{array}{l}\\ \text{a}.-7& & \text{d}.4\\ \text{b}.-5& & \text{e}.7\\ \text{c}.-4\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misalkan bahwa}f(x)=x^3+p{x}^2-3x+q\\ & \text{dengan}\\ & x^3+p{x}^2-3x+q=a{x}^3+b{x}^2+cx+d\\ & \text{Perhatikan bahwa}:\\ & f(-2)=(-2{)}^3+p(-2{)}^2-3(-2)+q=0\\ & \iff -8+4p+6+q=0\\ & \iff 4p+q=2......(1)\\ & f(3)=(3{)}^3+p(3{)}^2-3(3)+q=0\\ & \iff 27+9p-9+q=0\\ & \iff 9p+q=-18......(2)\\ & \text{Selanjutnya}\\ & \begin{array}{lllll}f(-2)& =& 4p+q& =& 2\\ f(3)& =& 9p+q& =& -18& -\\ & & -5p& =& 20\\ & & p& =& -4\\ & & \text{maka}q& =& 18\end{array}\\ & \text{Sehingga persamaan menjadi}\\ & x^3-4x^2-3x+18=0,\\ & \text{maka nilai}\\ & x_1+x_2+x_3=-{\displaystyle \frac{b}{a}=-\frac{(-4)}{1}=4}\end{array}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Nilai}m\text{agar-agar persamaan}\\ & x^3+3x^2-6x+m=0\text{membentuk}\\ & \text{barisan arirmetika adalah}....\\ & \begin{array}{l}\\ \text{a}.8& & \text{d}.-2\\ \text{b}.6& & \text{e}.-5\\ \text{c}.3\end{array}\\ \\ & \text{Jawab}:-\\ & \begin{array}{rl}& \text{Misalkan bahwa}f(x)=x^3+3x^2-6x+m\\ & \text{dengan}\\ & x^3+3x^2-6x+m=a{x}^3+b{x}^2+cx+d\\ & \text{Jika}{x}_1,x_2,x_3\text{akar-akarnya, maka}\\ & 2x_2=x_1+x_3\text{karena membentuk}\\ & \mathbf{\text{barisan aritmetika}}.\text{Selanjutnya}\\ & \bullet x_1+x_2+x_3={\displaystyle -\frac{b}{a}=-\frac{3}{1}=-3}\\ & \iff 2x_2+x_2=-3\iff x_2=-1\iff x_1+x_3=-2\\ & \bullet x_1{x}_2+x_1{x}_3+x_2{x}_3={\displaystyle \frac{c}{a}=\frac{-6}{1}=-6}\\ & \iff x_1+x_3+x_2{x}_3=-6\iff -2+x_2{x}_3=-6\\ & \iff x_2{x}_3=-4\iff (-1)x_3=-4\iff x_3=4\\ & \bullet x_1+x_3=-2\iff x_1+4=2\iff x_1=-2\\ & \bullet x_1{x}_2{x}_3=-{\displaystyle \frac{d}{a}=-\frac{m}{1}=-m}\\ & \iff m=-(x_1{x}_2{x}_3)=-(-2.-1.4)=-8\end{array}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Jika}\alpha ,\beta ,\gamma \text{merupakan akar persamaan}\\ & x^3-3x^2+4x+5=0\text{maka nilai}\\ & {\alpha }^3+{\beta }^3+{\gamma }^3\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.-30& & \text{d}.28\\ \text{b}.-28& & \text{e}.30\\ \text{c}.-24\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & x^3-3x^2+4x+5=a{x}^3+b{x}^2+cx+d\\ & \text{Karena memiliki akar-akar}:\alpha ,\beta ,\gamma ,\\ & \text{maka}\\ & \bullet {\alpha }^3-3{\alpha }^2+4\alpha +5=0\\ & \bullet {\beta }^3-3{\beta }^2+4\beta +5=0\\ & \bullet {\gamma }^3-3{\gamma }^2+4\gamma +5=0\\ & \text{Jika ketiganya dijumlahkan, diperoleh}\\ & {\alpha }^3+{\beta }^3+{\gamma }^3-3({\alpha }^2+{\beta }^2+{\gamma }^2)\\ & +4(\alpha +\beta +\gamma )+15=0\\ & \iff {\alpha }^3+{\beta }^3+{\gamma }^3=3({\alpha }^2+{\beta }^2+{\gamma }^2)\\ & -4(\alpha +\beta +\gamma )-15\\ & \iff {\alpha }^3+{\beta }^3+{\gamma }^3=3(\alpha +\beta +\gamma {)}^2\\ & -6(\alpha \beta +\alpha \gamma +\beta \gamma )-4(\alpha +\beta +\gamma )-15\\ & \iff {\alpha }^3+{\beta }^3+{\gamma }^3=3{(-{\displaystyle \frac{b}{a}})}^2-6\left({\displaystyle \frac{c}{a}}\right)\\ & -4(-{\displaystyle \frac{b}{a}})-15\\ & \iff {\alpha }^3+{\beta }^3+{\gamma }^3=3{(-(-3))}^2-6(4)\\ & -4(-(-4))-15\\ & \iff {\alpha }^3+{\beta }^3+{\gamma }^3=27-24-16-15=-28\end{array}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Jika}\alpha ,\beta ,\gamma \text{merupakan akar persamaan}\\ & x^3-14x^2+px+q=0\text{dengan}\\ & \alpha :\beta :\gamma =1:2:4,\text{maka nilai}p-q\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.160& & \text{d}.10\\ \text{b}.120& & \text{e}.8\\ \text{c}.100\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & x^3-14x^2+px+q=a{x}^3+b{x}^2+cx+d\\ & \text{Karena memiliki akar-akar}:\alpha ,\beta ,\gamma ,\\ & \text{dengan}\alpha :\beta :\gamma =1:2:4,\text{maka}\\ & \alpha =n,\beta =2n,\gamma =4n.\text{Perhatikan}\\ & \alpha +\beta +\gamma =-{\displaystyle \frac{b}{a}\iff n+2n+4n=14}\\ & \iff 7n=14\iff n=2.\text{Selanjutnya}\\ & \bullet \alpha =2\Rightarrow (2{)}^3-14(2{)}^2+p(2)+q=0\\ & \iff 8-56+2p+q=0\iff 2p+q=48\\ & \bullet \beta =4\Rightarrow (4{)}^3-14(4{)}^2+p(4)+q=0\\ & \iff 64-224+4p+q=0\iff 4p+q=160\\ & \text{Selanjutnya perhatikan eliminasi berikut}\\ & \begin{array}{llllr}4p& +& q& =& 160\\ 2p& +& q& =& 48& -\\ 2p& & & =& 112\\ & & q& =& -64& ,\text{maka}\\ p& -& q& =& 56& +64=120\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika diketahui}{x}^3-x-1=8\text{maka}\\ & x^4+x^3-x^2-2x+1=....\\ & \begin{array}{l}\\ \text{a}.0& & \text{d}.8x+8\\ \text{b}.2& & \text{e}.8x+10\\ \text{c}.8\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & x^3-x-1=8\text{saat dikali}x\\ & \text{masing-masing ruas, maka}\\ & x^4-x^2-x=8x.\text{Jika keduanya}\\ & \text{dijumlahkan}\\ & x^4+x^3-x^2-2x-1=8x+8\\ & \iff x^4+x^3-x^2-2x-1+2=8x+8+2\\ & \iff x^4+x^3-x^2-2x+1=8x+10\end{array}\end{array}$.

Contoh Soal Polinom (Bagian 6)

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

$\begin{array}{l}\\ 27.& \text{Salah satu akar dari polinomial}\\ & 2x^3+5x^2+x-2=0\text{adalah}{\displaystyle \frac{1}{2}}\\ & \text{Jumlah dua akar yang lain adalah}....\\ & \begin{array}{l}\\ \text{a}.6& & \text{d}.-3\\ \text{b}.3& & \text{e}.-6\\ \text{c}.2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|c|}\hline \begin{array}{rl}& \text{Perhatikan uraian berikut}\\ & {\displaystyle \frac{2x^3+5x^2+x-2}{(2x-1)}}\end{array}\\ \\ \begin{array}{lll}\mathbf{\text{pembagi}}& x^2+3x+2& \mathbf{\text{hasil}}\\ 2x-1& 2x^3+5x^2+x-2& \\ & 2x^3-x^2& -\\ & 6x^2+x-2& \\ & 6x^2-3x& -\\ & 4x-2\\ & 4x-2& -\\ \mathbf{\text{Sisa}}& 0& (\mathbf{\text{habis}})\end{array}\\ \\ \begin{array}{rl}\therefore f(x)& =2x^3+5x^2+x-2\\ & =(2x-1)(x^2+3x+2)\\ & =(2x-1)(x+1)(x+2)\end{array}\\ \hline\end{array}\\ & \text{Sehingga}{x}_2=-1,x_3=-2,\text{maka}\\ & x_2+x_3=-1+(-2)=-3\\ & \text{atau dapat juga ditentukan dengan rumus}\\ & \text{ABC pada persamaan kuadrat, yaitu}\\ & x^2+3x+2=a_2{x}^2+a_{1}x+a_0\\ & \iff x^2+3x+2=-{\displaystyle \frac{a_1}{a_1}=-{\displaystyle \frac{3}{1}=-3}}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Salah satu akar dari polinomial}\\ & x^4-5x^3+5x^2+5x-6=0\\ & \text{adalah}2.\text{Jumlah akar-akar yang }\\ & \text{lain adalah}....\\ & \begin{array}{l}\\ \text{a}.6& & \text{d}.3\\ \text{b}.5& & \text{e}.2\\ \text{c}.4\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|c|}\hline \begin{array}{rl}& \text{Perhatikan uraian berikut}\\ & {\displaystyle \frac{x^4-5x^3+5x^2+5x-6}{(x-2)}}\end{array}\\ \\ \begin{array}{lll}\mathbf{\text{pembagi}}& x^3-3x^2-x+3& \mathbf{\text{hasil}}\\ x-2& x^4-5x^3+5x^2+5x-6& \\ & x^4-2x^3& -\\ & -3x^3+5x^2+5x-6& \\ & -3x^3+6x^2& -\\ & -x^2+5x-6\\ & -x^2+2x& -\\ & 3x-6\\ & 3x-6& -\\ \mathbf{\text{Sisa}}& 0& (\mathbf{\text{habis}})\end{array}\\ \\ \begin{array}{rl}\therefore f(x)& =x^4-5x^3+5x^2+5x-6\\ & =(x-2)(x^3-3x^2-x+3)\end{array}\\ \hline\end{array}\\ & \text{Sehingga}\\ & x^3-3x^2-x+3=a_3{x}^3+a_2{x}^2+a_{1}x+a_0\\ & \iff x_1+x_2+x_3=-{\displaystyle \frac{a_2}{a_2}=-\frac{-3}{1}=3}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Akar-akar dari persamaan polinom}\\ & x^3-3x^2+4x+5=0\text{adalah}{x}_1,x_2\\ & \text{dan}{x}_3.\text{Nilai}{x}_1^2+x_2^2+x_3^2=....\\ & \begin{array}{l}\\ \text{a}.1& & \text{d}.17\\ \text{b}.3& & \text{e}.19\\ \text{c}.9\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& x^3-3x^2+4x+5=a{x}^3+b{x}^2+cx+d\\ & \text{Perhatikan bahwa}:\\ & x_1^2+x_2^2+x_3^2\\ & =(x_1+x_2+x_3{)}^2-2(x_1{x}_2+x_2{x}_3+x_1{x}_3)\\ & ={\left({\displaystyle \frac{-b}{a}}\right)}^2-2\left({\displaystyle \frac{c}{a}}\right)\\ & ={\left({\displaystyle \frac{-(-3)}{1}}\right)}^2-2\left({\displaystyle \frac{4}{1}}\right)=9-8=1\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Diketahui persamaan polinom}\\ & x^3-4x^2+6x-12=0\text{mempunyai}\\ & \text{akar-akar}{x}_1,x_2\text{dan}{x}_3.\\ & \text{Nilai}{\displaystyle \frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=....}\\ & \begin{array}{l}\\ \text{a}.-3& & \text{d}.{\displaystyle \frac{1}{2}}\\ \text{b}.-2& & \text{e}.3\\ \text{c}.{\displaystyle \frac{1}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& x^3-4x^2+6x-12=a{x}^3+b{x}^2+cx+d\\ & \text{Perhatikan bahwa}:\\ & {\displaystyle \frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}\\ & ={\displaystyle \frac{x_2{x}_3+x_1{x}_3+x_1{x}_2}{x_1{x}_2{x}_3}}\\ & ={\displaystyle \frac{{\displaystyle \frac{c}{a}}}{-{\displaystyle \frac{d}{a}}}=-\frac{c}{d}=-{\displaystyle \frac{6}{-12}=\frac{1}{2}}}\end{array}\end{array}$.

Contoh Soal Polinom (Bagian 5)

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

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

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

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

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

Ketaksamaan Minkowski

SELAMAT HARI RAYA IDUL FITRI 1443 H