Lanjutan Materi Polinom

Lanjutan Vektor di Ruang

Vektor di Ruang

 $\color{blue}\textrm{A. Vektor Di Ruang}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|c|c|}\hline \textrm{Nama}&\textbf{R}^{3}\\\hline \textrm{Vektor Satuan}&\textrm{Ruang (Bidang XYZ)}\\\hline \hat{e}_{\bar{a}}=\displaystyle \frac{\bar{a}}{\left | \bar{a} \right |}&\begin{cases} i= &\textrm{vektor satuan} \\ &\textrm{yang searah sumbu X}\\ j= &\textrm{Vektor satuan}\\ &\textrm{yang searah sumbu Y}\\ k=&\textrm{Vektor satuan}\\ &\textrm{searah sumbu Z} \end{cases} \\\hline \textrm{Vektor nol}&\overrightarrow{O}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\\\hline \textrm{Vektor posisi}&\overrightarrow{OP}=\vec{p}=\begin{pmatrix} p_{1}\\ p_{2}\\ p_{3} \end{pmatrix}=p_{1}\bar{i}+p_{2}\bar{j}+p_{3}\bar{j}\\\hline \textrm{Besar Vektor}&\overrightarrow{OP}=\sqrt{p_{1}^{2}+p_{2}^{2}+p_{3}^{2}}\\\hline \end{array}$

$\color{blue}\textrm{B. Operasi Vektor}$

$\color{blue}\textrm{1. Sifat-Sifat Aljabar Vektor}$

$\begin{array}{|l|l|l|}\hline 1.&\textrm{Komutatif penjumlahan}&\vec{a}+\vec{b}=\vec{b}+\vec{a}\\\hline 2.&\textrm{Asosiatif penjumlahan}&\left ( \vec{a}+\vec{b} \right )+\vec{c}=\vec{a}+\left ( \vec{b}+\vec{c} \right )\\\hline 3.&\textrm{Elemen Identitas}&\vec{a}+\vec{0}=\vec{0}+\vec{a}=\vec{a}\\\hline 4.&\textrm{Invers Penjumlahan}&\vec{a}+\left ( -\vec{a} \right )=\left ( -\vec{a} \right )+\vec{a}=\vec{0}\\\hline 5.&\textrm{Perkalian dengan skalar}&k\left ( l\vec{a} \right )=\left ( kl \right )\vec{a}\\ &&k\left ( \vec{a}+ \vec{b} \right )=k\vec{a}+k\vec{b}\\ &&k\left ( \vec{a}- \vec{b} \right )=k\vec{a}-k\vec{b}\\\hline 6.&\begin{aligned}&\textrm{Jika A, B, dan C segaris }\\ &\color{blue}\textrm{(Kolinear)} \end{aligned}&\begin{cases} \overrightarrow{AB}=k\overrightarrow{BC} \\ \overrightarrow{AC}=k\overrightarrow{AB} \\ dll \end{cases}\\\hline \end{array}$.

$\begin{array}{|c|c|}\hline \color{blue}\textrm{Vektor}&\color{blue}\textrm{Contoh}\\\hline \vec{z}=a\vec{i}+b\vec{j}+c\vec{k}&\begin{aligned}&\textrm{diketahui}\: \: \vec{p}=\vec{i}-2\vec{j}+2\vec{k}\\ &\textrm{maka pangjang vektor}\: \: \vec{p}\: \: \textrm{adalah}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &\quad\: \: =\sqrt{1+4+4}=\sqrt{9}=3 \end{aligned}\\\hline &\begin{aligned}&\textrm{Vektor satuan dari}\: \: \vec{p}\: \: \textrm{adalah}\\ &\vec{e}_{\vec{p}}=\frac{\vec{p}}{\left | \vec{p} \right |}=\displaystyle \frac{\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{3}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}=\displaystyle \begin{pmatrix} \frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3} \end{pmatrix} \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}+\overrightarrow{b}\\ &\textrm{b}.\quad 6\overrightarrow{a}+2\overrightarrow{b}\\ &\textrm{c}.\quad 2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}+\overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 2+(-3)\\ 1+(-5)\\ (-4)+2 \end{pmatrix}\\ &=\begin{pmatrix} 2-3\\ 1-5\\ -4+2 \end{pmatrix}=\color{red}\begin{pmatrix} -1\\ -4\\ -2 \end{pmatrix}\\ \textrm{b}.\quad&6\overrightarrow{a}+2\overrightarrow{b}=6\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+2\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ 6-10\\ -24+4 \end{pmatrix}=\color{red}\begin{pmatrix} 12\\ -4\\ -20 \end{pmatrix}\\ \textrm{c}.\quad&2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &2\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}-\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}+\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} 4+3+7\\ 2+5+0\\ -8-2+4 \end{pmatrix}=\color{red}\begin{pmatrix} 17\\ 7\\ -6 \end{pmatrix}\\ \textrm{d}.\quad&\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}=\cdots \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\textrm{tentukanlah}\: \: \left | \overrightarrow{a} \right |\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\left | \overrightarrow{a} \right |&=\sqrt{2^{2}+1^{2}+(-4)^{2}}\\ &=\sqrt{4+1+16}\\ &=\color{red}\sqrt{21} \end{aligned} \end{array}$

$\color{blue}\textrm{2. Perkalian Skalar Dua Vektor}$

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

$\color{red}\overrightarrow{a}\cdot \overrightarrow{b}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$.

Misalkan diketahui

$\color{red}\begin{aligned}&\overrightarrow{a}=\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}, \overrightarrow{b}=\begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix},\: \: \color{black}\textrm{maka}\\ & \overrightarrow{a} \cdot \overrightarrow{b} =\color{black}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}\cdot \begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix}\\ &\qquad\quad =\color{black}a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}\cdot \overrightarrow{b}\\ &\textrm{b}.\quad \overrightarrow{a}\cdot \overrightarrow{c}\\ &\textrm{c}.\quad \overrightarrow{b}\cdot \overrightarrow{c}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}\cdot \overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\cdot \begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=(2)(-3)+(1)(-5)+(-4)(2)\\ &=-6-5-8=\color{red}-19\\ \textrm{b}\quad&\overrightarrow{a}\cdot \overrightarrow{c}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\cdot \begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=(2)(7)+(1)(0)+(-4)(4)\\ &=14+0-16=\color{red}-2\\ \textrm{c}\quad&\overrightarrow{b}\cdot \overrightarrow{c}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\cdot \begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=(-3)(7)+(-5)(0)+(2)(4)\\ &=-21+0+8=\color{red}-13 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: t\: \: \textrm{jika}\\ & \overrightarrow{p}=3\bar{i}+t\bar{j}+\bar{k}\: \: \textrm{dan}\: \: \overrightarrow{p}\cdot \overrightarrow{p}=13\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\overrightarrow{p}\cdot \overrightarrow{p}=13\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |\left | \overrightarrow{p} \right |\cos 0^{\circ}=13,\\ &\qquad\qquad \color{blue}\textrm{ingat bahwa}\: \: \angle \left ( \overrightarrow{p},\overrightarrow{p} \right )=0^{\circ}\\ &\qquad\qquad \color{blue}\textrm{dan nilai}\: \: \cos 0^{\circ}=1,\\ & \color{black}\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |^{2}.1=13\Leftrightarrow \left | \overrightarrow{p} \right |^{2}=13\\ &\Leftrightarrow \left (\sqrt{3^{2}+t^{2}+1^{2}} \right )^{2}=13\\ &\Leftrightarrow 3^{2}+t^{2}+1^{2}=13\\ &\Leftrightarrow 9+t^{2}+1=13\\ &\Leftrightarrow t^{2}=13-9-1=10\\ &\Leftrightarrow t^{2}=3\\ &\Leftrightarrow t=\color{red}\pm \sqrt{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: \overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{Jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q}\: \: \textrm{maka}\\ &\textrm{tentukanlah nilai}\: \: t\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{dengan}\: \: \overrightarrow{p}\: \: \textrm{dan}\: \: \overrightarrow{q}\: \: \textrm{tegak lurus}\\ &\textrm{artinya}\: \: \color{blue}\angle \left ( \overrightarrow{p},\overrightarrow{q} \right )=90^{\circ}.\: \color{black}\textrm{Sehingga}\\ &\textrm{nilai}\: \: \color{blue}\cos 90^{\circ}=0\\ &\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cos \theta \\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cdot 0=0\\ &\Leftrightarrow \: \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\cdot \begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ &\Leftrightarrow \: (-2)(4)+(1)(-1)+(3)(t)=0\\ &\Leftrightarrow \: -8-1+3t=0\\ &\Leftrightarrow \: 3t=9\\ &\Leftrightarrow \: t=\color{red}3 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Polinom

 $\color{blue}\textrm{A. Pendahuluan}$

Polinom disebut juga suku banyak. Polinom atau suku banyak adalah suatu bentuk variabel yang berpangkat/berderajat.

Secara definisi suku banyak (polinomial) dalam  $x$  berderajat $n$ adalah:

Suatu bentuk

$\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}$

dengan  $n$  bilangan cacah serta  $a_{0},\: a_{1},\: a_{2},\: ...,\: a_{n}$  koefisien dari suku  $x$  dan  $a_{n}\neq 0$  dengan  $a_{0}$  sebagai suku tetap (konstan)nya.

Selanjutnya perhatikanlah tabel berikut!

$\color{red}\begin{array}{|l|l|}\hline \begin{aligned}a_{n}&\: \: \textrm{adalah koefisien dari} \: \: x^{n}\\ a_{n-1}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-1}\\ a_{n-2}&\: \: \textrm{adalah koefisien dari} \: \: x^{n-2}\\ \vdots &\\ a_{2}&\: \: \textrm{adalah koefisien dari} \: \: x^{2}\\ a_{1}&\: \: \textrm{adalah koefisien dari} \: \: x^{1}\\ a_{0}&\: \: \textrm{adalah konstanta} \\ &(\textrm{suku tetap}) \end{aligned}&\begin{aligned}a_{n}\: &\: \neq 0\\ n:&\: \: \textrm{bilangan cacah},\\ :&\: \: \textrm{adalah derajat (pangkat)} \\ &\: \: \textrm{tertinggi dalam suku} \\ &\: \: \textrm{banyak tersebut}&\\ &\\ &\\ &\end{aligned}\\\hline \end{array}$

$\LARGE\colorbox{yellow}{CONTOH SOAL 1}$

$\begin{aligned}1.\quad&\textrm{Polinom}\: \: \color{red}2x^{3}-6x^{2}+2020\: \: \color{black}\textrm{dapat dinyatakan}\\ &\textrm{dengan}\: \: \: \color{blue}2x^{3}-6x^{2}+0x^{1}+2020x^{0}\\ &\textrm{Polinom tersebut memiliki suku tetap}\: \: 2020\\ 2.\quad&\textrm{Polinom}\: \: \color{red}5x^{4}-8x^{3}+6x-2021 \: \: \color{black}\textrm{dapat dinyatakan}\\ &\textrm{dengan}\: \: \: \color{blue}5x^{4}-8x^{3}+0x^{2}+6x^{1}-2021x^{0}\\ &\textrm{Polinom tersebut memiliki suku tetap}\: \: -2021\\ 3.\quad&\textrm{Polinom}\: \: \color{red}x^{4}-2x^{3}+3x^{2}-2\sqrt{x}+1 \: \: \color{black}\textrm{tidak dapat}\\ &\textrm{dinamakan polinom, sebab ada variabel dari}\: \: \: \color{blue}x\\ &\textrm{yang berderajat bukan bilangan cacah}\\ 4.\quad&\textrm{Sedangkan polinom}\: \: \color{red}5-x+(2-x)(1+x+x^{2})\\ &\textrm{adalah bentuk polinom, karena dapat dinayatakan}\\ &\textrm{dengan}\: \: \: \color{blue}-x^{3}+x^{2}+7 \end{aligned}$

$\color{blue}\textrm{B. Nilai Polinom}$

Polinom atau suku banyak yang berderajat $\color{red}n$ yang selanjutnya dinyatakan dengan 

$f(x)=\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{1}x^{1}+a_{0}$

Berkaitan dengan kebutuhan penentuan nilai ini, dapat ditentukan dengan dua cara:

$\textbf{a. Substitusi}$

$\begin{aligned}&\textrm{Nilai suku banyak}\: \: \color{red}f(x)\: \: \textrm{berderajat}\\ &n\: \: \textrm{saat}\: \: \color{red}x = k\: \: \color{black}\textrm{adalah}\: \: \color{blue}f(k).\\ &\textrm{Jika}\: \: f(k)=0\: \: \textrm{maka}\: \: x = k\: \: \textrm{akar dari}\: \: f(x),\\ &\textrm{dan}\: \: (x-k)\: \: \textrm{faktor dari}\: \: f(x)\\ &\end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL 2}$

Jika suatu polinom dinyatakan dengan  $f(x)$, maka nilai polinom itu untuk  $x=3$  adalah  $f(3)$.

Misalkan diketahui  

$\begin{aligned}1.\quad f(x)&=x^{3}-1\\ \textrm{mak}&\textrm{a}\\ f(1)&=1^{3}-1=1-1=0\\ f(3)&=3^{3}-1=27-1=26\\ f(-4)&=(-4)^{2}-1=-64-1=-65 \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: h(x)=2x^{3}+5x^{2}-12x-6\\ &\textrm{Tentukanlah nilai untuk}\: \: h(-2),\: h(-1),\\ &h(0),\: h(1),\: \: \textrm{dan}\: \: h(2)\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{|c|c|l|}\hline \color{red}x=k&\color{red}h(k)&\qquad\qquad\qquad\qquad\color{red}\textrm{Nilai}\\\hline x=-2&h(-2)&\begin{aligned}h(-2)&=2(-2)^{3}+5(-2)^{2}-12(-2)-6\\ &=-16+20+24-6\\ &=22 \end{aligned}\\\hline x=-1&h(-1)&\begin{aligned}h(-1)&=2(-1)^{3}+5(-1)^{2}-12(-1)-6\\ &=-2+5+12-6\\ &=9 \end{aligned}\\\hline x=0&h(0)&\begin{aligned}h(0)&=2(0)^{3}+5(0)^{2}-12(0)-6\\ &=-6 \end{aligned}\\\hline x=1&h(1)&\begin{aligned}h(1)&=2(1)^{3}+5(1)^{2}-12(1)-6\\ &=2+5-12-6\\ &=-11 \end{aligned}\\\hline x=2&h(2)&\begin{aligned}h(2)&=2(2)^{3}+5(2)^{2}-12(2)-6\\ &=16+20-24-6\\ &=6 \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: p(x)=x-2019\\ &\textrm{dan}\: \: q(x)=x^{2019}+1.\: \textrm{Tentukanlah}\\ &\textrm{nilai untuk}\: \: p\left ( q(2) \right )\: \: \textrm{dan}\: \: q\left ( p(2) \right )\\\\ &\color{blue}\textrm{Jawab}:\\ &\textrm{Yang dibahas yang bagian}\: \: p\left ( q(2) \right )\\ &q(2)=2^{2019}+1,\: \textrm{maka nilai}\\ &\begin{aligned}p\left ( q(2) \right )&=\left ( 2^{2019}+1 \right )-2019\\ &=2^{2019}-2018 \end{aligned}\\\\ &\textrm{Untuk yang}\: \: q\left ( p(2) \right )\: \: \textrm{adalah}\\ &p(2)=\cdots , \: \textrm{maka nilai}\\ &\begin{aligned}q\left ( p(2) \right )&=\because \cdots ^{2019}+1\\ &=\cdots \end{aligned} \end{array}$

$\textbf{b. Horner/Sintetik}$

Nilai suatu polinom dapat ditentukan dengan pembagian sintesis Horner

Misalkan:

$\begin{aligned}f(x)&=\color{blue}ax^{3}+bx^{2}+cx+d\: \: \color{black}\textrm{saat akan dibagi}\\ &\color{red}x=h,\: \: \color{black}\textrm{maka pembagian Horner itu}:\\ & \end{aligned}$

Perhatikan bahwa proses ke bawah adalah berup proses penjumlahan.

Proses di atas akan sama saat kita mensubstitusikan  $\color{red}x=h$  ke dalam  $\color{red}f(x)$, yaitu:

$\begin{aligned}f(x)&=\color{blue}ax^{3}+bx^{2}+cx+d\: \: \textrm{saat}\\ &\color{red}x=h,\: \: \color{black}\textrm{maka}\\ f(\color{red}h\color{black})&=a\color{red}h^{3}\color{black}+b\color{red}h^{2}\color{black}+c\color{red}h\color{black}+d\\ &\\ &\textbf{Cukup JELAS bukan}? \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL 3}$

$\begin{array}{l}\\ \textrm{Tentukanlah nilai dari}\: \: f(4)\: \: \textrm{jika}\\ \textrm{diketahui}\: \: f(x)=x^{3}-x-5\\ \textrm{Jawab}:\\ \begin{aligned}(1).\quad&\textrm{Cara substitusi langsung}\\ &f(x)=x^{3}-x-5\\ &f(4)=\color{red}4^{3}-4-5\\ &\qquad=\color{red}64-9=\color{blue}55\\ (2).\quad&\textrm{Cara Horner}\\ &\textrm{Karena}\: \: f(x)=x^{3}-x-5\\ &\textrm{dan koefisiennya yang akan}\\ &\textrm{adalah}:\\ & a_{3}=1,\: a_{2}=0,\: a_{1}=-1,\: \&\: a_{0}=-5\\ &\textbf{maka bagan pembagian Hornernya}\\ &\begin{array}{ll|llllllllll}\\ &\color{red}x=4&1&\color{blue}0&\color{magenta}-1&-5&\\ &&&&&&\\ &&&\color{blue}4&\color{magenta}16&60&+\\\hline &&1&\color{blue}4&\color{magenta}15&55 \end{array} \end{aligned} \end{array}$