Contoh Soal Polinom (Bagian 9)

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

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)

$\begin{array}{ll}\\ 36.&(\textbf{Soal Seleksi OM Af-Sel 1983})\\ &\textrm{Jika diketahui}\: \: a^{3}-a-1=0\: \: \textrm{maka}\\ & a^{4}+a^{3}-a^{2}-2a+1\: =\:  ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0&&\textrm{d}.\quad 1\\ \textrm{b}.\quad  a+1&\qquad&\textrm{e}.\quad a^{3}+a+1\\ \textrm{c}.\quad  \color{red}2\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Lihat pembahasan no}.35\\ &\textrm{Cukup jelas }   \end{aligned}   \end{array}$ .

$\begin{array}{ll}\\ 37.&\textrm{Tentukanlah suku banyak}\: \: f(x)\: \: \textrm{sedemikian}\\ &\textrm{sehingga}\: \: f(x)\: \: \textrm{terbagi oleh}\: \: x^{2}+1,\\ &\textrm{sedangkan}\: \: f(x)+1\: \: \textrm{terbagi oleh}\: \: x^{3}+x^{2}+1\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\left ( x^{2}+1 \right ).h_{1}x\\ f(x)+1&=\left ( x^{2}+1 \right ).h_{1}x+1\\ \textrm{supaya}\: \: \: &f(x)+1\: \: \textrm{terbagi habis oleh}\: \: x^{3}+x^{2}+1 ,\\ & \textrm{maka akan ada bilangan bulat}\: \: k,\: \: \left ( k\neq 0 \right )\\ k&=\displaystyle \frac{f(x)+1}{x^{3}+x^{2}+1}\\ &=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\\ k=1\Rightarrow &1=\displaystyle \frac{\left ( x^{2}+1 \right ).h_{1}x+1}{x^{3}+x^{2}+1}\: \: \textrm{maka}\: \: h_{1}x=x\\ \textrm{sehingga}&\: \: f(x)=\color{red}x^{3}+x^{2}\\ \textrm{untuk ni}&\textrm{lai}\: \: k\: \: \textrm{yang lain, tak ditemukan} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\color{blue}(\textrm{KSM 2015})\color{black}\textrm{Diketahui}\: \: f(x)\: \: \textrm{adalah polinom}\\ & (x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})(x-x_{5})\\ &\textrm{dengan}\: \: \: x_{1},\: x_{2},\: x_{3},\: x_{4},\: \: \textrm{dan}\: \: x_{5}\: \: \textrm{adalah}\\ &\textrm{bilangan bulat berbeda}.\: \textrm{Jika}\: \: f(104)=2012,\\ &\textrm{maka nilai} \: \: \: x_{1}+ x_{2}+ x_{3}+ x_{4}+x_{5}\: \: \textrm{sama dengan}....\\ &\textrm{a}.\quad 13\\ &\textrm{b}.\quad 14\\ &\textrm{c}.\quad 16\\ &\textrm{d}.\quad \color{red}17\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}:\\ f(x)&=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})(x-x_{5})\\ f(104)&=(104-x_{1})(104-x_{2})(104-x_{3})(104-x_{4})(104-x_{5})=2012\\ &=2012=1\times 2\times 503\\ &=(-1)\times (1)\times (-2)\times (2)\times (503)\\ \textrm{maka}&\: \: \begin{cases} (104-x_{1}) &=-2\Rightarrow x_{1}=106 \\ (104-x_{2}) &=-1\Rightarrow x_{2}=105 \\ (104-x_{3}) &=1\Rightarrow x_{3}=103 \\ (104-x_{4}) &=2\Rightarrow x_{4}=102 \\ (104-x_{5}) &=503\Rightarrow x_{5}=-399 \\ \end{cases}\\ \textrm{sehin}&\textrm{gga},\\ &x_{1}+ x_{2}+ x_{3}+ x_{4}+x_{5}=106+105+103+102+(-399)=\color{red}17 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Akar-akar persamaan}\\ &(x+1)(2x+1)(3x-1)(4x-1)+6x^{4}=0\\ &\textrm{adalah}\: \:  x_{1},x_{2},x_{3}\: \: \textrm{dan}\: \: x_{4}\: .\: \textrm{Jika}\\ &x_{1}<x_{2}<x_{3}< x_{4}\: \: \textrm{dan}\: \: x_{1}+x_{4}=m\\ &\textrm{serta}\: \: x_{2}+x_{3}=n,\: \textrm{maka}\: \: mn=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{7}{30}&&\textrm{d}.\quad \color{red}\displaystyle \frac{2}{15}\\ \textrm{b}.\quad  \displaystyle \frac{7}{30}&\qquad&\textrm{e}.\quad -\displaystyle \frac{4}{15}\\ \textrm{c}.\quad  \displaystyle \frac{4}{15}\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &(x+1)(2x+1)(3x-1)(4x-1)+6x^{4}=0\\ &\Leftrightarrow (2x^{2}+3x+1)(12x^{2}-7x+1)+6x^{4}=0\\ &\Leftrightarrow 24x^{4}+22x^{3}-7x^{2}-4x+1+6x^{4}=0\\ &\Leftrightarrow 30x^{4}+22x^{3}-7x^{2}-4x+1+6x^{4}=0\\ &\Leftrightarrow (6x^{2}+2x+2)(5x^{2}+2x-1)=0\\ &\Leftrightarrow (6x^{2}+2x+2)=0\: \textrm{V}\: (5x^{2}+2x-1)=0\\ &\begin{cases} x_{1} &=\displaystyle \frac{-1-\sqrt{6}}{5} \\  x_{2} &=\displaystyle \frac{-1-\sqrt{7}}{6} \end{cases},\: \begin{cases} x_{3} &=\displaystyle \frac{-1+\sqrt{7}}{6} \\  x_{4} &=\displaystyle \frac{-1+\sqrt{6}}{5}  \end{cases}\\ &\textrm{Dengan}\: \: m=x_{1}+x_{4}=-\displaystyle \frac{2}{5},\: n=x_{2}+x_{3}=-\displaystyle \frac{1}{3}\\ &\textrm{maka}\: \: mn=\left ( -\displaystyle \frac{2}{5} \right )\left ( -\displaystyle \frac{1}{3} \right )=\color{red}\displaystyle \frac{2}{15}  \end{aligned}    \end{array}$ 

$\begin{array}{ll}\\ 40.&\textrm{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^{2}+x+1=0\\ & \textrm{adalah}\: \: x_{1},\: x_{2},\: x_{3},...,x_{2017}\\ &\textrm{Tentukan nilai dari}\\ & \displaystyle \frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\frac{1}{1-x_{3}}+...+\displaystyle \frac{1}{1-x_{2017}}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \frac{x^{2018}-1}{x-1}&=x^{2017}+x^{2016}+x^{2015}+...+x^{2}+x+1=0\\ \textrm{perlu dii}&\textrm{ngat bahwa kondisi ini mensyaratkan}\: \: x\neq 1,\\ & \textrm{sehingga}\\\\ x^{2018}-1&=0\\ x^{2018}&=1\\ x&=\pm 1,\: \: \: \: \textrm{pilih}\: \: x=-1\\ \textrm{maka}\: \quad &\textrm{nilai dari}\\ \displaystyle \frac{1}{1-x_{1}}+&\frac{1}{1-x_{2}}+\frac{1}{1-x_{3}}+...+\displaystyle \frac{1}{1-x_{2017}}\\ &=\underset{\textrm{sebanyak 2017}}{\underbrace{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}\\ &=\underset{\textrm{sebanyak 2017}}{\underbrace{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+\displaystyle \frac{1}{2}}} \\ &=\color{red}\displaystyle \frac{2017}{2} \end{aligned} \end{array}$.

Contoh Soal Polinom (Bagian 7)

$\begin{array}{ll}\\ 31.&\textrm{Diketahui faktor-faktor polinom}\\ & x^{3}+px^{2}-3x+q=0\: \: \textrm{adalah}\\ &(x+2)\: \: \textrm{dan}\: \: (x-3)\: .\: \textrm{Jika akar-akar}\\  &\textrm{polinom tersebut adalah}\: \:  x_{1},x_{2}\: \: \textrm{dan}\: \: x_{3},\\ &\textrm{maka nilai}\: \: x_{1}+x_{2}+x_{3}=\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -7&&\textrm{d}.\quad \color{red}4\\ \textrm{b}.\quad  -5&\qquad&\textrm{e}.\quad 7\\ \textrm{c}.\quad  -4\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Misalkan bahwa}\: \:  f(x)=x^{3}+px^{2}-3x+q\\ &\textrm{dengan}\\ &x^{3}+px^{2}-3x+q=ax^{3}+bx^{2}+cx+d\\ &\textrm{Perhatikan bahwa}:\\ &f(-2)=(-2)^{3}+p(-2)^{2}-3(-2)+q=0\\ &\: \:  \quad\quad \Leftrightarrow -8+4p+6+q=0\\ &\: \:  \quad\quad \Leftrightarrow  4p+q=\color{red}2\: \color{black}......(1)\\ &f(3)=(3)^{3}+p(3)^{2}-3(3)+q=0\\ &  \quad\quad \Leftrightarrow 27+9p-9+q=0\\ &  \quad\quad \Leftrightarrow  9p+q=\color{red}-18\: \color{black}......(2)\\ &\textrm{Selanjutnya}\\ &\begin{array}{llllll} f(-2)&=&4p+q&=&2\\ f(3)&=&9p+q&=&-18&-\\\hline &&-5p&=&20\\ &&\qquad p&=&-4\\ &&\textrm{maka}\: \: q&=&18  \end{array}\\ &\textrm{Sehingga persamaan menjadi}\\ &x^{3}-4x^{2}-3x+18=0,\\ &\textrm{maka nilai}\\ &x_{1}+x_{2}+x_{3}=-\displaystyle \frac{b}{a}=-\frac{(-4)}{1}=4    \end{aligned}   \end{array}$.

$\begin{array}{ll}\\ 32.&\textrm{Nilai}\: \: m\: \: \textrm{agar-agar persamaan}\\ & x^{3}+3x^{2}-6x+m=0\: \: \textrm{membentuk}\\ &\textrm{barisan arirmetika adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 8&&\textrm{d}.\quad -2\\ \textrm{b}.\quad  6&\qquad&\textrm{e}.\quad -5\\ \textrm{c}.\quad  3\\  \end{array}\\\\ &\textrm{Jawab}:-\\ &\begin{aligned}&\textrm{Misalkan bahwa}\: \:  f(x)=x^{3}+3x^{2}-6x+m\\ &\textrm{dengan}\\ &x^{3}+3x^{2}-6x+m=ax^{3}+bx^{2}+cx+d\\ &\textrm{Jika}\: \: x_{1},x_{2},x_{3}\: \: \textrm{akar-akarnya, maka}\\ &2x_{2}=x_{1}+x_{3}\: \: \textrm{karena membentuk}\\ &\textbf{barisan aritmetika}\: .\: \textrm{Selanjutnya}\\ &\bullet \: \: x_{1}+x_{2}+x_{3}=\displaystyle -\frac{b}{a}=-\frac{3}{1}=-3\\ &\Leftrightarrow 2x_{2}+x_{2}=-3\Leftrightarrow x_{2}=-1\Leftrightarrow x_{1}+x_{3}=-2\\ &\bullet \: \: x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=\displaystyle \frac{c}{a}=\frac{-6}{1}=-6\\ &\Leftrightarrow x_{1}+x_{3}+x_{2}x_{3}=-6\Leftrightarrow -2+x_{2}x_{3}=-6\\ &\Leftrightarrow x_{2}x_{3}=-4\Leftrightarrow (-1)x_{3}=-4\Leftrightarrow x_{3}=4\\ &\bullet x_{1}+x_{3}=-2\Leftrightarrow x_{1}+4=2\Leftrightarrow x_{1}=-2\\ &\bullet x_{1}x_{2}x_{3}=-\displaystyle \frac{d}{a}=-\frac{m}{1}=-m\\ &\Leftrightarrow m=-(x_{1}x_{2}x_{3})=-(-2.-1.4)=-8     \end{aligned}   \end{array}$.

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

$\begin{array}{ll}\\ 34.&\textrm{Jika}\: \: \alpha ,\beta ,\gamma \: \: \textrm{merupakan akar persamaan}\\ & x^{3}-14x^{2}+px+q=0\: \: \textrm{dengan}\\ &\alpha :\beta :\gamma =1:2:4\: ,\: \textrm{maka nilai}\: \: p-q\\ &\textrm{adalah}\: ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 160&&\textrm{d}.\quad 10\\ \textrm{b}.\quad  \color{red}120&\qquad&\textrm{e}.\quad 8\\ \textrm{c}.\quad  100\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &x^{3}-14x^{2}+px+q=ax^{3}+bx^{2}+cx+d\\ &\textrm{Karena memiliki akar-akar}\: :\: \alpha ,\beta ,\gamma ,\\  &\textrm{dengan}\: \: \alpha :\beta :\gamma =1:2:4,\: \textrm{maka}\\ &\alpha =n,\: \beta =2n,\: \gamma =4n\: .\: \textrm{Perhatikan}\\ &\alpha +\beta +\gamma =-\displaystyle \frac{b}{a}\Leftrightarrow n+2n+4n=14\\ &\Leftrightarrow 7n=14\Leftrightarrow n=2\: .\: \color{red}\textrm{Selanjutnya}\\ &\bullet \quad \alpha =2\Rightarrow  (2)^{3}-14(2)^{2}+p(2)+q=0\\ &\Leftrightarrow  8-56+2p+q=0\Leftrightarrow \color{blue}2p+q=48\\ &\bullet \quad \beta  =4\Rightarrow  (4)^{3}-14(4)^{2}+p(4)+q=0\\ &\Leftrightarrow  64-224+4p+q=0\Leftrightarrow \color{blue}4p+q=160\\ &\textrm{Selanjutnya perhatikan eliminasi berikut}\\ &\begin{array}{llllrl} 4p&+&q&=&160\\ 2p&+&q&=&48&-\\\hline 2p&&&=&112\\ &&q&=&-64&,\: \textrm{maka}\\ p&-&q&=&56&+\: 64=\color{red}120   \end{array}   \end{aligned}    \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Jika diketahui}\: \: x^{3}-x-1=8\: \: \textrm{maka}\\ & x^{4}+x^{3}-x^{2}-2x+1\: =\:  ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 0&&\textrm{d}.\quad 8x+8\\ \textrm{b}.\quad  2&\qquad&\textrm{e}.\quad \color{red}8x+10\\ \textrm{c}.\quad  8\\  \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &x^{3}-x-1=8\: \: \textrm{saat dikali}\: \: x\\ &\textrm{masing-masing ruas, maka}\\ &x^{4}-x^{2}-x=8x\: .\: \textrm{Jika keduanya}\\ &\textrm{dijumlahkan}\\ &x^{4}+x^{3}-x^{2}-2x-1=8x+8\\ &\Leftrightarrow x^{4}+x^{3}-x^{2}-2x-1+2=8x+8+2\\ &\Leftrightarrow x^{4}+x^{3}-x^{2}-2x+1=\color{red}8x+10   \end{aligned}   \end{array}$.

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