Contoh Soal Suku Banyak

\begin{array}{ll}\\ \fbox{1}.&\textrm{Jika}\: \: g(x)=2x^{3}+x^{2}-x+1,\: \: \textrm{maka}\: \: g(1)=\: \: \\ &\begin{array}{lll}\\ \textrm{a}.\quad -2&&\textrm{d}.\quad 2\\ \textrm{b}.\quad -1&\textrm{c}.\quad 1&\textrm{e}.\quad 3 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}g(x)&=2x^{3}+x^{2}-x+1\\ g(1)&=2(1)^{3}+(1)^{2}-(1)+1\\ &=2+1-1+1\\ &=3 \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Jika}\: \: p(y)=5y^{4}+2r^{2}y^{3}+y^{2}+1\: \: \textrm{dan}\: \: q(y)=4y^{5}+3ry^{2}-3y-1\: \: \\ &\textrm{serta}\: \: p(-1)=q(-1),\: \: \textrm{maka nilai}\: \: r\: \: \textrm{sama dengan}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{2}\: \: \textrm{dan}\: \: 3&&\textrm{d}.\quad -\displaystyle \frac{3}{2}\\ \textrm{b}.\quad \displaystyle -\frac{3}{2}\: \: \textrm{dan}\: \: 3&\textrm{c}.\quad \displaystyle \frac{3}{2}\: \: \textrm{dan}\: \: -3&\textrm{e}.\quad 3 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{c}\\\\ \begin{aligned}p(-1)&=q(-1)\\ 5(-1)^{4}+2r^{2}(-1)^{3}+(-1)^{2}+1&=4(-1)^{5}+3r(-1)^{2}-3(-1)-1\\ 5-2r^{2}+1+1&=-4+3r+3-1\\ 9-3r-2r^{2}&=0\\ \displaystyle \frac{(-6-2r)(-3+2r)}{2}&=0,\qquad \textrm{ingat pemfaktoran}\\ (-3-r)(-3+2r)&=0\\ r=-3\quad \vee \quad r&=\displaystyle \frac{3}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui}\: \: f(x)\: \: \textrm{berderajat}\: \: n. \: \: \textrm{Jika pembaginya berbentuk}\: \: \left ( ax^{2}+bx+c \right ),\\ &\textrm{dengan}\: \: a\neq 0,\: \: \textrm{maka hasil baginya berderajat}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad n-1&&\textrm{d}.\quad 3\\ \textrm{b}.\quad n-2&\textrm{c}.\quad n-3&\textrm{e}.\quad 2 \end{array}\end{array}\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}\textrm{Suku banyak (polinom)}&=\textrm{pembagi}\times \textrm{hasil bagi}+\textrm{sisa}\\ x^{n}+...&=\left ( ax^{2}+bx+c \right )\times \left ( x^{n-2}+... \right )+\left (mx+n \right ) \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Hasil bagi dan sisanya jika}\: \: \left (6x^{4}-3x^{2}+x-1 \right )\: \: \textrm{dibagi oleh}\: \: \left ( 2x-1 \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+\frac{1}{8}&\textrm{dan}&-\displaystyle \frac{7}{8}\\ \textrm{b}.\quad 3x^{3}+3x^{2}-\displaystyle \frac{3}{4}x+1&\textrm{dan}&-7\\ \textrm{c}.\quad x^{3}+\displaystyle \frac{3}{2}x^{2}-3x+\frac{1}{8}&\textrm{dan}&\displaystyle \frac{7}{8}\\ \textrm{d}.\quad x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+1&\textrm{dan}&\displaystyle \frac{1}{8}\\ \textrm{e}.\quad 3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x-\frac{1}{8}&\textrm{dan}&-\displaystyle \frac{7}{8}\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{array}{rr|rrrrrrr} x=\frac{1}{2}&&6&0&-3&1&-1\\ &&&3&\frac{3}{2}&-\frac{3}{4}&\frac{1}{8}&+&\\\cline{3-7} \multicolumn{3}{r}{6}&3&-\frac{3}{2}&\frac{1}{4}&\boxed{-\frac{7}{8}} \end{array}\begin{cases} \textrm{Hasil bagi}: & \displaystyle \frac{6x^{3}+3x^{2}-\frac{3}{2}x+\frac{1}{4}}{2}=3x^{3}+\displaystyle \frac{3}{2}x^{2}-\displaystyle \frac{3}{4}x+\displaystyle \frac{1}{8} \\ & \\ \textrm{Sisa bagi}: & -\displaystyle \frac{7}{8} \end{cases}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Hasil bagi dan sisanya jika}\: \: \left (x^{4}-x^{3}-x^{2}+x-1 \right )\: \: \textrm{dibagi oleh}\: \: \left ( x-2 \right )\left ( x+1 \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{2}+1&\textrm{dan}&2x+1\\ \textrm{b}.\quad x^{2}+1&\textrm{dan}&2x-1\\ \textrm{c}.\quad x^{2}-1&\textrm{dan}&2x+1\\ \textrm{d}.\quad x^{2}-1&\textrm{dan}&2x-1\\ \textrm{e}.\quad 2x^{2}-1&\textrm{dan}&x+1\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \textrm{Dengan menggunakan pembagian}\: \: \textbf{Horner-Kino}\: \: \textrm{akan diperoleh},\\\\ \begin{array}{rr|rrr|rrr} &&1&-1&-1&1&-1\\ 2&&\ast &\ast &2&0&2&\\\cline{3-7} 1&&\ast &1&0&1&\ast &+\\\cline{3-7} \multicolumn{3}{r}{1}&0&1&2&1\\\cline{6-7} \end{array}\quad \Rightarrow \begin{cases} \textrm{Suku banyak}: & f(x)=x^{4}-x^{3}-x^{2}+x-1 \\ \textrm{Pembagai}: & p(x)=(x-2)(x+1)=x^{2}-x-2 \\ &: 2\: \: \textrm{dari}\: -\frac{-2}{1},\: \: \textrm{sedang}\: \: 1=-\left ( \frac{-1}{1} \right )\\ \textrm{Hasil bagi}:&h(x)=x^{2}+1\\ \textrm{Sisa bagi}:&s(x)=2x+1 \end{cases} \\\\ \textrm{Sehingga},\\\\ x^{4}-x^{3}-x^{2}+x-1=\left ( x^{2}-x-2 \right )\left ( x^{2}+1 \right )+2x+1.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Diketahui bahwa}\: \: \displaystyle \frac{f(x)}{x-2}=h(x)+\displaystyle \frac{3}{x-2}\: \: \textrm{dan}\: \: \displaystyle \frac{f(x)}{x-1}=h(x)+\displaystyle \frac{2}{x-1}\: ,\\ &\textrm{jika}\: \: \displaystyle \frac{f(x)}{(x-2)(x-1)}=h(x)+\displaystyle \frac{s(x)}{(x-2)(x-1)}\: ,\: \: \textrm{maka}\: \: s(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x+1&&\textrm{d}.\quad 2x-1\\ \textrm{b}.\quad x+2&\textrm{c}.\quad 2x+1&\textrm{e}.\quad x-2\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}\displaystyle \frac{f(x)}{x-2}=h(x)+\displaystyle \frac{3}{x-2}\Rightarrow f(x)&=(x-2).h(x)+3\Rightarrow f(2)=3\\ \displaystyle \frac{f(x)}{x-1}=h(x)+\displaystyle \frac{2}{x-1}\Rightarrow f(x)&=(x-1).h(x)+2\Rightarrow f(1)=2\\ \displaystyle \frac{f(x)}{(x-2)(x-1)}&=h(x)+\displaystyle \frac{s(x)}{(x-2)(x-1)}\\ \textrm{maka}\qquad\qquad f(x)&=(x-2)(x-1).h(x)+s(x)\\ f(x)&=(x-2)(x-1).h(x)+px+q\\ f(2)&=2p+q=3\\ f(1)&=p+q=2,\\ \textrm{sehingga dengan }&\textrm{eliminasi akan diperoleh}\\ p&=1\quad \textrm{dan}\\ q&=1\\ \textrm{Jadi},\quad\qquad px+q&=x+1 \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: x^{4}+2mx-n\: \: \textrm{dibagi}\: \: x^{2}-1\: \: \textrm{bersisa}\: \: 2x-1\: ,\\ &\textrm{maka nilai}\: \: m\: \: \textrm{dan}\: \: n\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad m=-1\: \: \textrm{dan}\: \: n=2&&\textrm{d}.\quad m=-1\: \: \textrm{dan}\: \: n=-2\\ \textrm{b}.\quad m=1\: \: \textrm{dan}\: \: n=-2&\textrm{c}.\quad m=1\: \: \textrm{dan}\: \: n=2&\textrm{e}.\quad m=-2\: \: \textrm{dan}\: \: n=1\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{c}\\\\ \textrm{Dengan menggunakan pembagian}\: \: \textbf{Horner-Kino}\: \: \textrm{akan diperoleh},\\\\ \begin{array}{rr|rrr|rrr} &&1&0&0&2m&-n\\ 1&&\ast &\ast &1&0&1&\\\cline{3-7} 0&&\ast &0&0&0&\ast &+\\\cline{3-7} \multicolumn{3}{r}{1}&0&1&2m&1-n\\\cline{6-7} \end{array}\quad \Rightarrow \begin{cases} \textrm{Suku banyak}: & f(x)=x^{4}+2mx-n \\ \textrm{Pembagai}: & p(x)=(x-1)(x+1)=x^{2}-1 \\ &: 1\: \: \textrm{dari}\: -\frac{-1}{1},\: \: \textrm{sedang}\: \: 0=-\left ( \frac{0}{1} \right )\\ \textrm{Hasil bagi}:&h(x)=x^{2}+1\\ \textrm{Sisa bagi}:&s(x)=2mx+(1-n)=2x-1 \end{cases} \\\\ \textrm{Sehingga},\\\\ 2m=2\Rightarrow m=1\\ 1-n=-1\Rightarrow n=2.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: f(x)=x^{4}-kx^{2}+5\: \: \textrm{habis dibagi}\: \: (x-1)\: \: , \textrm{maka}\: \: f(x)\: \: \textrm{juga habis dibagi oleh}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x+1&&\textrm{d}.\quad x+5\\ \textrm{b}.\quad 2x+1&\textrm{c}.\quad 3x+1&\textrm{e}.\quad 2x+5 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}f(x)&=x^{4}-kx^{2}+5\\ f(1)&=(1)^{4}-k(1)^{2}+5\\ 0&=1-k+5\\ k&=6\\ f(x)&=x^{4}-6x^{2}+5\\ &=(x^{2}-1)(x^{2}-5)\\ &=(x-1)(x+1)(x^{2}-5) \end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: (m-2)\: \: \textrm{adalah faktor dari}\: \: 2m^{3}+3tm+4\: \: , \textrm{maka nilai}\: \: t\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{10}{3}&&\textrm{d}.\quad -\displaystyle \frac{3}{10}\\ \textrm{b}.\quad \displaystyle \frac{1}{3}&\textrm{c}.\quad \displaystyle \frac{3}{10}&\textrm{e}.\quad -\displaystyle \frac{10}{3} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}f(m)&=2m^{3}+3tm+4\\ f(2)&=2(2)^{3}+3t(2)+4\\ 0&=16+6t+4\\ -6t&=20\\ t&=-\displaystyle \frac{10}{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{(KSM MA Kab/Kota 2015)Nilai terkecil}\: \: n\: \: \textrm{yang mengkin sehingga}\: \: n.(n+1).(n+2)\\\ & \textrm{habis dibagi 24 adalah}....\\ &\begin{array}{l}\\ \textrm{a}.\quad 1\\ \textrm{b}.\quad 2\\ \textrm{c}.\quad 3\\ \textrm{d}.\quad 4 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}k&=\displaystyle \frac{n.(n+1).(n+2)}{24}\\ &=\displaystyle \frac{n.(n+1).(n+2)}{2.(2+1).(2+2)}\\ \textrm{m}&\textrm{aka}\: \: n=2 \end{aligned}.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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