Exercise Zone : Teorema Sisa (Remainder Theorem)

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Tipe:

Standar SBMPTN HOTS

No. 1

Diketahui suku banyak {P(x) = 2x^4 + ax^3-3x^2 + 5x + b}. Jika P(x) dibagi {(x − 1)} sisa 11, dibagi {(x + 1)} sisa -1, maka nilai {(2a + b) =}

1. 13
2. 10
   
3. 8
   

4. 7
5. 6
   

GANESHA OPERATION

$\begin{array}{rl}P(1)& =11\\ 2(1{)}^4+a(1{)}^3-3(1{)}^2+5(1)+b& =11\\ 2+a-3+5+b& =11\\ a+b& =7\end{array}$

$\begin{array}{rl}P(-1)& =-1\\ 2(-1{)}^4+a(-1{)}^3-3(-1{)}^2+5(-1)+b& =-1\\ 2-a-3-5+b& =-1\\ -a+b& =5\end{array}$

CARA BIASA	CARA CEPAT
$\begin{array}{rl}a+b& =7\\ -a+b& =5& +\\ 2b& =12\\ b& =6\end{array}$ $\begin{array}{rl}a+b& =7\\ a+6& =7\\ a& =1\end{array}$ $\begin{array}{rl}2a+b& =2(1)+6\\ & =2+6\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$	$\begin{array}{rlr}a+b& =7& \times 3\\ -a+b& =5& \times -1\end{array}$ $\begin{array}{rl}3a+3b& =21\\ a-b& =-5& +\\ 4a+2b& =16\\ 2a+b& =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, 2a+b=8.
JAWAB: C

No. 2

Jika f(x) dibagi {(x-3)} sisanya 5 sedangkan jika dibagi {(x+1)} sisanya 1. Jika f(x) dibagi dengan {x^2-2x-3} sisanya adalah

1. {x+2}
2. {2x+3}
   
3. {3x-1}
   

4. {2x+1}
5. {x+3}
   

f(3)=5
f(-1)=1
Misal f(x) dibagi x^2-2x-3 sisanya adalah ax+b

x^2-2x-3(x-3)(x+1)

f(3)=3a+b=5
f(-1)=-a+b=1

$\begin{array}{rl}3a+b& =5\\ -a+b& =1& -\\ 4a& =4\\ a& =1\end{array}$

$\begin{array}{rl}-a+b& =1\\ -1+b& =1\\ b& =2\end{array}$ JAWAB: A

No. 3

Jika {f(x)=ax^3+2bx^2-bx+2} dibagi dengan {(x-1)} sisanya 3, sedangkan jika dibagi dengan {(x-2)} sisanya -4. Nilai {2a+b} adalah

1. 5
2. -5
   
3. -7
   

4. -12
5. 1
   

$\begin{array}{rl}f(1)& =3\\ a(1{)}^3+2b(1{)}^2-b(1)+2& =3\\ a+2b-b+2& =3\\ a+b& =1\end{array}$

$\begin{array}{rl}f(2)& =-4\\ a(2{)}^3+2b(2{)}^2-b(2)+2& =-4\\ 8a+8b-2b+2& =-4\\ 8a+6b& =-6\\ 4a+3b& =-3\end{array}$

$\begin{array}{rlr}a+b& =1& \times 3\\ 4a+3b& =-3\end{array}$

$\begin{array}{rl}3a+3b& =3\\ 4a+3b& =-3-\\ -a& =6\\ a& =-6\end{array}$

$\begin{array}{rl}a+b& =1\\ -6+b& =1\\ b& =7\end{array}$

$\begin{array}{rl}2a+b& =2(-6)+7\\ & =-12+7\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -5}}}}\end{array}$

Jadi, 2a+b=-5.
JAWAB: B

No. 4

Jika {f(x)=ax^3+3bx^2+(2a-b)x+4} dibagi dengan {(x-1)} sisanya 10, sedangkan jika dibagi dengan {(x+2)} sisanya 2. Nilai {a+b} adalah

1. 1
2. \dfrac43
   
3. \dfrac73
   

4. 2
5. 3
   

$\begin{array}{rl}f(1)& =10\\ a(1{)}^3+3b(1{)}^2+(2a-b)(1)+4& =10\\ a+3b+2a-b+4& =10\\ 3a+2b& =6\end{array}$

$\begin{array}{rl}f(-2)& =2\\ a(-2{)}^3+3b(-2{)}^2+(2a-b)(-2)+4& =2\\ -8a+12b-4a+2b+4& =2\\ -12a+14b& =-2\\ 6a-7b& =1\end{array}$

$\begin{array}{rlr}3a+2b& =6& \times 2\\ 6a-7b& =1\end{array}$

$$\begin{array}{rl}6a+4b& =12\\ 6a-7b& =1-\\ 11b& =11\\ b& =1\end{array}$$\)

$\begin{array}{rl}3a+2b+b& =6+1\\ 3a+3b& =7\\ a+b& =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{7}{3}}}}}}\end{array}$

Jadi, a+b=\dfrac73.
JAWAB: C

No. 5

Suku banyak f(x) dibagi {(x+1)} sisanya -2 dan dibagi {(x-3)} sisa 7, suku banyak g(x) dibagi {(x+1)} sisa 3 dan dibagi {(x-3)} sisa 2. Diketahui {h(x)=f(x)\cdot g(x)}, jika h(x) dibagi \left(x^2-2x-3\right) sisanya adalah

1. S(x)=3x-1
2. S(x)=4x-1
   
3. S(x)=5x-1
   

4. S(x)=6x-1
5. S(x)=7x+2
   

f(-1)=-2
f(3)=7
g(-1)=3
g(3)=2

Misal sisanya adalah ax+b

x^2-2x-3=(x+1)(x-3)

$$\begin{array}{rl}h(-1)=a(-1)+b& =f(-1)g(-1)\\ -a+b& =(-2)(3)\\ -a+b& =-6\end{array}$$

$$\begin{array}{rl}h(3)=a(3)+b& =f(3)g(3)\\ 3a+b& =(7)(2)\\ 3a+b& =14\end{array}$$

$$\begin{array}{rl}3a+b& =14\\ -a+b& =-6& -\\ 4a& =20\\ a& =5\end{array}$$

$$\begin{array}{rl}-a+b& =-6\\ -5+b& =-6\\ b& =-1\end{array}$$

Jadi, sisanya adalah S(x)=5x-1.
JAWAB: C

No. 6

Jika suku banyak f(x) dibagi oleh x-2 menghasilkan sisa 10, sisa pembagian suku banyak f(x) oleh {x^2-3x+2} adalah...

1. {f(1)(2-x)-10(x-1)}
2. {f(1)(x-2)+10(x-1)}
   
3. {f(1)(x-2)-10(x+1)}
   

4. {f(1)(2-x)+10(x-1)}
5. {f(1)(2-x)-10(x+1)}
   

f(2)=10.

x^2-3x+2=(x-1)(x-2)

Sisanya adalah,
$$\begin{array}{rl}S(x)& ={\displaystyle \frac{x-a}{b-a}}f(b)+{\displaystyle \frac{x-b}{a-b}}f(a)\\ & ={\displaystyle \frac{x-1}{2-1}}f(2)+{\displaystyle \frac{x-2}{1-2}}f(1)\\ & ={\displaystyle \frac{x-1}{1}}10+{\displaystyle \frac{x-2}{-1}}f(1)\\ & =10(x-1)+f(1)(2-x)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle f(1)(2-x)+10(x-1)}}}}\end{array}$$

Jadi, sisa pembagian suku banyak f(x) oleh {x^2-3x+2} adalah {f(1)(2-x)+10(x-1)}.
JAWAB: D