Matematika Dasar Suku Banyak Atau Polinomial (*Soal Dari Banyak Sekali Sumber)

pekerjaan siswa tampaknya masih jauh dari apa yang diharapkan7. SBMPTN 2015
Sisa pembagian $x^{2014}-Ax^{2015}+Bx^{3}-1$ oleh $x^{2}-1$ yaitu $–x+B$. Nilai $2A+B$ adalah...
A. $1$
B. $2$
C. $3$
D. $4$
E. $5$Alternatif Pembahasan:

Hint

Untuk menuntaskan soal diatas, kita coba mengingatkan kembali perihal teorema sisa, yaitu:
Untuk
$F(x)=H(x)\cdot P(x)+Sisa$
$F(x)=H(x)\cdot (x-a)(x-b)+mx+n$
maka
$F(a)=am+n$
$F(b)=bm+n$

Pada soal disampaikan bahwa $x^{2014}-Ax^{2015}+Bx^{3}-1$ dibagi oleh $x^{2}-1$ sisanya $-x+B$.
$x^{2014}-Ax^{2015}+Bx^{3}-1=\left (x^{2}-1 \right )\cdot H(x)+sisa$
$x^{2014}-Ax^{2015}+Bx^{3}-1=\left (x-1 \right )\left (x+1 \right )\cdot H(x)-x+B$
$x^{2014}-Ax^{2015}+Bx^{3}-1=\left (x-1 \right )\left (x+1 \right )\cdot H(x)-x+B$

Untuk $x=1$
$1-A+B-1=-1+B$
$-A=-1$
$A=1$

Untuk $x=-1$
$-1+A-B-1=1+B$
$A-B=1+B$
$1-B=1+B$
$B=0$

$\therefore 2A+B=2$ $\B$

8. UM UNDIP 2015
Jika suku banyak $f(x)$ dibagi dengan $(x-a)(x-b)$ dengan $a \neq b$, maka sisa pembagian ini adalah...
A. $\frac{x+a}{a-b}f\left ( a \right )+\frac{x+b}{b-a}f\left ( b \right )$
B. $\frac{x-a}{a-b}f\left ( b \right )+\frac{x-b}{b-a}f\left ( a \right )$
C. $\frac{x+a}{a-b}f\left ( b \right )+\frac{x+b}{b-a}f\left ( a \right )$
D. $\frac{x-b}{a-b}f\left ( a \right )+\frac{x-a}{b-a}f\left ( b \right )$
E. $\frac{x-b}{a-b}f\left ( b \right )+\frac{x-a}{b-a}f\left ( a \right )$Alternatif Pembahasan:

Hint

Untuk menuntaskan soal diatas, kita coba memakai teoerma sisa, yaitu:
Untuk
$f(x)=h(x)\cdot (x-a)(x-b)+mx+n$
maka
$f(a)=am+n$
$f(b)=bm+n$

Kita terapkan ke soal dengan mengeliminasi $n$ atau mengeliminasi $m$;
#mengeliminasi $n$
$f(a)-f(b)=am-bm$
$f(a)-f(b)=( a-b)m$
$m=\frac{f(a)-f(b)}{a-b}$

#mengeliminasi $m$
$b \cdot f(a)-a \cdot f(b)=bn-an$
$b \cdot f(a)-a \cdot f(b)=( b-a)n$
$n=\frac{b \cdot f(a)-a \cdot f(b)}{b-a}$

$\therefore $ Sisa Pembagian yaitu $mx+n$
$mx+n=\frac{f(a)-f(b)}{a-b}x+\frac{b \cdot f(a)-a \cdot f(b)}{b-a}$
$mx+n=\frac{f(a)-f(b)}{a-b}x+\frac{a \cdot f(b)-b \cdot f(a)}{a-b}$
$mx+n=\frac{x \cdot f(a)-x \cdot f(b)+a \cdot f(b)-b \cdot f(a)}{a-b}$
$mx+n=\frac{x \cdot f(a)-b \cdot f(a)-x \cdot f(b)+a \cdot f(b)}{a-b}$
$mx+n=\frac{x-b}{a-b}f(a)+\frac{a-x}{a-b}f(b)$
$mx+n=\frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)$ $\D$

9. SBMPTN 2014
Diketahui $P(x)$ suatu polynomial. Jika $P(x+1)$ dan $P(x-1)$ masing-masing memperlihatkan sisa $2$ apabila masing-masing dibagi $x-1$,
maka $P(x)$ dibagi $x^{2}-2x$ memperlihatkan sisa...
A. $x+2$
B. $2x$
C. $x$
D. $1$
E. $2$Alternatif Pembahasan:

Hint

$P(x)=H(x)\cdot (x-a)(x-b)+mx+n$

Untuk $x=0$
$P(x)=H(x)\cdot x(x-2)+mx+n$
maka $P(0)=n$

Untuk $x=2$
$P(x)=H(x)\cdot x(x-2)+mx+n$
$P(2)=2m+n$

Pada soal diketahui $P(x+1)=2$ dan $P(x-1)=2$ maka untuk $x=1$ diperoleh $P(2)=2$ dan $P(0)=2$.

$P(0)=2$ dan $P(0)=n$ maka $n=2$
$P(2)=2$ dan $P(2)=2m+n$ maka $2m+n=2$, $m=0$

$\therefore $ Sisa pembagian yaitu $mx+n$ yaitu $2$ $\E$

10. SBMPTN 2016
Diketahui sisa pembagian suku banyak $f(x)-2g(x)$, oleh $x^{2}+x-2$ yaitu $x+3$, sisa pembagian $2f(x)+g(x)$ oleh $x^{2}-3x+2$ yaitu $x+1$, maka sisa pembagian $f(x)g(x)$ oleh $x-1$ adalah...
A. $\frac{23}{24}$
B. $\frac{18}{24}$
C. $-\frac{21}{25}$
D. $-\frac{48}{25}$
E. $-\frac{50}{36}$Alternatif Pembahasan:

Hint

Dari keterangan pada soal kita peroleh;
$f(x)-2g(x)=(x^{2}+x-2)H(x)+x+3$
$f(x)-2g(x)=(x+2)(x-1)H(x)+x+3$

$2f(x)+g(x)=(x^{2}-3x+2)H(x)+x+1$
$2f(x)+g(x)=(x-2)(x-1)H(x)+x+1$

Untuk $x=1$, kita peroleh;
$f(1)-2g(1)=4$ $\cdots (1)$
$2f(1)+g(1)=1$ $\cdots (2)$

jikalau pers $(1)$ dan $(2)$ kita eliminasi atau substitusi kita peroleh;
$g(1)=-\frac{6}{5}$ dan $f(1)=\frac{8}{5}$

$\therefore f(1)g(1)=\frac{8}{5}\frac{-6}{5}=-\frac{48}{25}$ $\D$

11. UN 2011
Diketahui suku banyak $P(x)=2x^{4}+ax^{3}-3x^{2}+5x+b$. Jika $P(x)$ dibagi $(x-1)$ sisa $11$ dan dibagi $(x+1)$ sisa $-1$, maka nilai $(2a+b)$ adalah...
A. $13$
B. $10$
C. $8$
D. $7$
E. $6$Alternatif Pembahasan:

Hint

Dari soal kita peroleh beberapa data, antara lain;
Jika $P(x)$ dibagi $(x-1)$ sisa $11$ maka $P(1)=11$
Jika $P(x)$ dibagi $(x+1)$ sisa $-1$ maka $P(-1)=-1$

Karena $P(1)=11$ maka
$P(x)=2x^{4}+ax^{3}-3x^{2}+5x+b$
$P(1)=2+a-3+5+b$
$11=a+b+4$
$a+b=7$ $\cdots (1)$

Karena $P(-1)=-1$ maka
$P(x)=2x^{4}+ax^{3}-3x^{2}+5x+b$
$P(-1)=2-a-3-5+b$
$-1=-a+b-64$
$-a+b=5$ $\cdots (2)$

Dengan mengeliminasi atau substitusi pers. $(1)$ dan $(2)$, maka kita peroleh nilai $a=1$ dan $b=6$

$\therefore 2a+b=2+6=8$ $\C$

12. UN 2007
Suku banyak $f(x)$ dibagi $(x+1)$ sisanya $10$ dan jikalau dibagi $(2x-3)$ sisanya $5$. Jika suku banyak $f(x)$ dibagi $(2x^{2}-x-3)$, sisanya adalah...
A. $-2x+8$
B. $-2x+12$
C. $-x+4$
D. $-5x+5$
E. $-5x+15$Alternatif Pembahasan:

Hint

Dari apa yang disampaikan pada soal, ada beberapa hal yang sanggup kita simpulkan yaitu;
$f(-1)=10$ dan $f(\frac{3}{2})=5$

Dari bentuk suku banyak;
$f(x)=h(x)\cdot p(x)+sisa$
$f(x)=h(x)\cdot 2x^{2}-x-3+mx+n$
$f(x)=h(x)\cdot (x+1)(2x-3)+mx+n$

$f(-1)=-m+n$ maka $-m+n=10$ $\cdots (1)$
$f(\frac{3}{2})=\frac{3}{2}m+n$ maka $\frac{3}{2}m+n=5$ $\cdots (2)$

Dengan mengeliminasi atau substitusi pers. $(1)$ dan $(2)$ kita peroleh nilai $m=-2$ atau $n=8$

$\therefore mx+n \equiv -2x+8$ $\A$

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