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Jika dibagi sisanya 20, jika dibagi sisanya 14. Jika  dibagi , sisanya adalah...

Pertanyaan

Jika f left parenthesis x right parenthesis dibagi left parenthesis x plus 2 right parenthesis sisanya 20, jika dibagi left parenthesis x plus 8 right parenthesis sisanya 14. Jika f left parenthesis x right parenthesis dibagi x squared plus 10 x plus 16, sisanya adalah...

  1. x minus 22

  2. x plus 22

  3. x minus 18

  4. x plus 18

  5. x minus 25

Pembahasan Soal:

Ingat teorema sisa:

Jika f left parenthesis x right parenthesis dibagi oleh x squared plus 10 x plus 16, maka:

f open parentheses x close parentheses equals H open parentheses x close parentheses open parentheses x plus 2 close parentheses open parentheses x plus 8 close parentheses plus S open parentheses x close parentheses 

misalkan S open parentheses x close parentheses equals a x plus b

Sehingga,

f open parentheses x close parentheses equals H open parentheses x close parentheses open parentheses x plus 2 close parentheses open parentheses x plus 8 close parentheses plus a x plus b

Untuk x equals negative 2

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell H open parentheses x close parentheses open parentheses x plus 2 close parentheses open parentheses x plus 8 close parentheses plus a x plus b end cell row cell f open parentheses negative 2 close parentheses end cell equals cell H open parentheses negative 2 close parentheses open parentheses open parentheses negative 2 close parentheses plus 2 close parentheses open parentheses open parentheses negative 2 close parentheses plus 8 close parentheses plus a open parentheses negative 2 close parentheses plus b end cell row 20 equals cell negative 2 a plus b... open parentheses 1 close parentheses end cell end table 

Untuk x equals negative 4

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell H open parentheses x close parentheses open parentheses x plus 2 close parentheses open parentheses x plus 8 close parentheses plus a x plus b end cell row cell f open parentheses negative 8 close parentheses end cell equals cell H open parentheses negative 8 close parentheses open parentheses open parentheses negative 8 close parentheses plus 2 close parentheses open parentheses open parentheses negative 8 close parentheses plus 8 close parentheses plus a open parentheses negative 8 close parentheses plus b end cell row 14 equals cell negative 8 a plus b... open parentheses 2 close parentheses end cell row blank blank blank end table

kemudian eliminasi persamaan (1) dan (2):

table row cell 20 equals negative 2 a plus b end cell row cell 14 equals negative 8 a plus b end cell row cell 6 equals 6 a end cell row cell a equals 1 end cell end table minus

Subtitusi nilai a ke persamaan (1):

20 equals negative 2 a plus b 20 equals negative 2 open parentheses 1 close parentheses plus b b equals 20 plus 2 b equals 22

Sehingga, jika f left parenthesis x right parenthesis dibagi oleh x squared plus 10 x plus 16 bersisa x plus 22 

Jadi, jawaban yang tepat adalah B

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Nasrullah

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Dengan menggunakan cara Horner, tentukan hasil bagi dan sisa pembagian suku banyak  oleh !

Pembahasan Soal:

Diketahui suku banyak 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 dibagi oleh 2 x squared minus 5 x plus 2. Faktor dari 2 x squared minus 5 x plus 2 sebagai berikut.

2 x squared minus 5 x plus 2 equals open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses

Dengan menggunakan cara horner, pembagian suku banyak 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 oleh open parentheses 2 x minus 1 close parentheses sebagai berikut.

long division with stack on the left by table row 4 0 cell negative 2 end cell 8 10 6 row blank 2 1 cell negative 1 half end cell cell 15 over 4 end cell cell 55 over 8 end cell end table yields table row 4 2 cell negative 1 end cell cell space 15 over 2 end cell cell 55 over 4 end cell cell left enclose bottom enclose 103 over 8 end enclose end enclose end cell end table pile 1 half end pile end long division

Berdasarkan permbagian dengan horner di atas, sisa pembagian suku banyak 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 oleh open parentheses 2 x minus 1 close parentheses adalah 103 over 8. Lalu, berdasarkan teorema sisa, f open parentheses 1 half close parentheses equals 103 over 8.

Kemudian, dengan menggunakan cara horner juga, pembagian suku banyak 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 oleh  open parentheses x minus 2 close parentheses sebagai berikut.

long division with stack on the left by table row 4 row blank end table space table row 0 row 8 end table space table row cell negative 2 end cell row 16 end table space table row 8 row 28 end table space table row 10 row 72 end table space table row 6 row 164 end table subscript plus yields table row 4 end table space table row 8 end table space table row cell space 14 end cell end table space table row 36 end table space table row 82 end table space table row cell bottom enclose left enclose 170 end enclose end cell end table pile 2 end pile end long division

Berdasarkan permbagian dengan horner di atas, sisa pembagian suku banyak 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 oleh open parentheses x minus 2 close parentheses adalah 170. Lalu, berdasarkan teorema sisa, f open parentheses 2 close parentheses equals 170.

Selanjutnya, misal f open parentheses x close parentheses equals 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6h open parentheses x close parentheses adalah hasil pembagian suku banyak f open parentheses x close parentheses oleh 2 x squared minus 5 x plus 2 dan s open parentheses x close parentheses adalah sisanya, maka dapat dituliskan:

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell open parentheses 2 x squared minus 5 x plus 2 close parentheses times h open parentheses x close parentheses plus s open parentheses x close parentheses end cell row cell f open parentheses x close parentheses end cell equals cell open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses times h open parentheses x close parentheses plus s open parentheses x close parentheses end cell end table

Misal s open parentheses x close parentheses equals a x plus b, maka diperoleh persamaan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses times h open parentheses x close parentheses plus a x plus b end cell row cell f open parentheses 1 half close parentheses end cell equals cell open parentheses 2 open parentheses 1 half close parentheses minus 1 close parentheses open parentheses 1 half minus 2 close parentheses times h open parentheses x close parentheses plus a open parentheses 1 half close parentheses plus b end cell row cell 103 over 8 end cell equals cell 0 times open parentheses negative 3 over 2 close parentheses times h open parentheses x close parentheses plus 1 half a plus b end cell row cell 1 half a plus b end cell equals cell 103 over 8 end cell row b equals cell 103 over 8 minus 1 half a space space space left parenthesis Persamaan space 1 right parenthesis end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses times h open parentheses x close parentheses plus a x plus b end cell row cell f open parentheses 2 close parentheses end cell equals cell open parentheses 2 open parentheses 2 close parentheses minus 1 close parentheses open parentheses 2 minus 2 close parentheses times h open parentheses x close parentheses plus 2 a plus b end cell row 170 equals cell 3 times 0 times h open parentheses x close parentheses plus 2 a plus b end cell row cell 2 a plus b end cell equals 170 row b equals cell 170 minus 2 a space space space left parenthesis Persamaan space 2 right parenthesis end cell end table
 

Kemudian, dengan menggunakan metode substitusi (substitusi persamaan 1 pada persamaan 2) diperoleh nilai a yang memenuhi sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row b equals cell 170 minus 2 a end cell row cell 103 over 8 minus 1 half a end cell equals cell 170 minus 2 a end cell row cell negative 1 half a plus 2 a end cell equals cell 170 minus 103 over 8 end cell row cell 3 over 2 a end cell equals cell fraction numerator 1.360 over denominator 8 end fraction minus 103 over 8 end cell row cell 3 over 2 a end cell equals cell fraction numerator 1.257 over denominator 8 end fraction end cell row a equals cell fraction numerator 2 cross times 1.257 over denominator 3 cross times 8 end fraction end cell row blank equals cell fraction numerator up diagonal strike 2 to the power of 1 cross times 1.257 over denominator up diagonal strike 24 subscript 12 end fraction end cell row blank equals cell fraction numerator 1.257 over denominator 12 end fraction end cell row blank equals cell 419 over 4 end cell end table

Lalu, substitusi a equals 419 over 4 pada persamaan 2, sehingga diperoleh nilai b yang memenuhi sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row b equals cell 170 minus 2 a end cell row blank equals cell 170 minus 2 open parentheses 419 over 4 close parentheses end cell row blank equals cell 170 minus up diagonal strike 2 open parentheses 419 over up diagonal strike 4 subscript 2 close parentheses end cell row blank equals cell 340 over 2 minus 419 over 2 end cell row blank equals cell negative 79 over 2 end cell end table

Berdasarkan nilai a dan b di atas, sisa pembagian f open parentheses x close parentheses oleh 2 x squared minus 5 x plus 2 sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell s open parentheses x close parentheses end cell equals cell a x plus b end cell row blank equals cell 419 over 4 x minus 79 over 2 end cell end table

Lalu, hasil pembagiannya sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses times h open parentheses x close parentheses plus s open parentheses x close parentheses end cell row cell f open parentheses x close parentheses minus s open parentheses x close parentheses end cell equals cell open parentheses 2 x squared minus 5 x plus 2 close parentheses times h open parentheses x close parentheses end cell row cell open parentheses 2 x squared minus 5 x plus 2 close parentheses times h open parentheses x close parentheses end cell equals cell f open parentheses x close parentheses minus s open parentheses x close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 x squared minus 5 x plus 2 close parentheses times h open parentheses x close parentheses end cell row blank equals cell 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 minus open parentheses 419 over 4 x minus 79 over 2 close parentheses end cell row blank equals cell 4 x to the power of 5 minus 2 x cubed plus 8 x squared minus 379 over 4 x plus 91 over 2 end cell row cell h open parentheses x close parentheses end cell equals cell fraction numerator 4 x to the power of 5 minus 2 x cubed plus 8 x squared minus 379 over 4 x plus 91 over 2 over denominator open parentheses 2 x squared minus 5 x plus 2 close parentheses end fraction end cell end table

Nilai h open parentheses x close parentheses di atas akan susah apabila ditentukan dengan cara horner, maka dengan menggunakan pembagian bersusun, nilai h open parentheses x close parentheses sebagai berikut.

long division by 2 x squared minus 5 x plus 2 yields table row cell 2 x cubed end cell cell plus 5 x squared end cell cell plus 19 over 2 x end cell cell plus 91 over 4 end cell blank end table pile table row cell 4 x to the power of 5 end cell cell negative 2 x cubed end cell cell plus 8 x squared end cell cell negative 379 over 4 x end cell cell plus 91 over 2 end cell blank end table bottom enclose table row cell space 4 x to the power of 5 end cell cell negative 10 x to the power of 4 end cell cell plus 4 x cubed space space space space space space space space space space space space space space space space space space space space end cell cell blank subscript minus end cell end table end enclose end pile end long division space space space space space space space space space space space space space space space space space space space space space space space table row cell 10 x to the power of 4 end cell cell negative 6 x cubed end cell cell plus 8 x squared end cell cell negative 379 over 4 x end cell cell plus 91 over 2 end cell blank end table space space space space space space space space space space space space space space space space space space space space space space space bottom enclose table row cell 10 x to the power of 4 end cell cell negative 25 x cubed end cell cell plus 10 x squared end cell end table space space space space space space space space space space space space space space space space space space space end enclose subscript minus space space space space space space space space space space space space space space space space space space space space space space table row cell 19 x cubed end cell cell negative 2 x squared end cell cell negative 379 over 4 x end cell cell plus 91 over 2 end cell blank blank end table space space space space space space space space space space space space space space space space space space space space space space space bottom enclose table row cell 19 x cubed end cell cell negative 95 over 2 x squared end cell cell plus 19 x end cell end table space space space space space space space space space space space space space space space space space space space end enclose subscript minus space space space space space space space space space space space space space space space space space space space space space space table row cell 91 over 2 x squared end cell cell negative 455 over 4 x end cell cell plus 91 over 2 end cell blank blank blank end table space space space space space space space space space space space space space space space space space space space space space space space bottom enclose table row cell 91 over 2 x squared end cell cell negative 455 over 4 x end cell cell plus 19 over 2 end cell end table space space space space space space space space space space space space space space space space space space space end enclose subscript minus space space space space space space space space space space space space space space space space space space space space space space space space space space 0


Berdasarkan pembagian di atas, h open parentheses x close parentheses equals 2 x cubed plus 5 x squared plus 19 over 2 x plus 91 over 4.

Dengan demikian, hasil pembagian suku banyak 4 x to the power of 5 minus 2 x cubed plus 8 x squared plus 10 x plus 6 oleh 2 x squared minus 5 x plus 2 adalah h open parentheses x close parentheses equals 2 x cubed plus 5 x squared plus 19 over 2 x plus 91 over 4 dan sisa pembagiannya adalah 419 over x minus 79 over 2.

Roboguru

Polinomial  habis dibagi . Polinomial  dibagi  bersisa . Jika  dibagi oleh , sisa pembagiannya adalah ...

Pembahasan Soal:

Jika begin mathsize 14px style p open parentheses x close parentheses end style habis dibagi begin mathsize 14px style open parentheses x squared minus 16 close parentheses equals open parentheses x minus 4 close parentheses open parentheses x plus 4 close parentheses end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p open parentheses 4 close parentheses end cell equals 0 row cell p open parentheses negative 4 close parentheses end cell equals 0 end table end style

Jika begin mathsize 14px style p open parentheses x close parentheses end style dibagi begin mathsize 14px style open parentheses x squared minus 9 close parentheses equals open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses end style bersisa begin mathsize 14px style open parentheses 5 x minus 2 close parentheses end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p open parentheses 3 close parentheses end cell equals cell 5 times 3 minus 2 equals 13 end cell row cell p open parentheses negative 3 close parentheses end cell equals cell 5 open parentheses negative 3 close parentheses minus 2 equals negative 17 end cell end table end style

Jika begin mathsize 14px style p open parentheses x close parentheses end style dibagi oleh begin mathsize 14px style open parentheses x squared plus 7 x plus 12 close parentheses end style, maka sisanya maksimum berderajat 1, yaitu begin mathsize 14px style S open parentheses x close parentheses equals a x plus b end style

begin mathsize 14px style p open parentheses x close parentheses equals open parentheses x plus 4 close parentheses open parentheses x plus 3 close parentheses times H open parentheses x close parentheses plus a x plus b end style

Substitusi begin mathsize 14px style x equals negative 4 end style dan begin mathsize 14px style x equals negative 3 end style diperoleh

begin mathsize 14px style p open parentheses negative 4 close parentheses equals negative 4 a plus b equals 0 p open parentheses negative 3 close parentheses equals negative 3 a plus b equals negative 17 end style

Dengan menggunakan metode eliminasi dan substitusi diperoleh begin mathsize 14px style a equals negative 17 end style dan begin mathsize 14px style b equals negative 68 end style

Sisa pembagian fungsi tersebut adalah begin mathsize 14px style S open parentheses x close parentheses equals a x plus b equals negative 17 x minus 68 end style 

Oleh karena itu, jawaban yang tepat adalah A.

Roboguru

Jika sisa pembagian polinomial  oleh  adalah , sisa pembagian polinomial  oleh  adalah ....

Pembahasan Soal:

Ingat kembali Teorema Sisa:

straight p open parentheses straight x close parentheses equals Pembagi times Hasil space bagi plus sisa straight p open parentheses straight x close parentheses equals Pembagi times straight H open parentheses straight x close parentheses plus straight S open parentheses straight x close parentheses 

Diketahui:

straight p open parentheses straight x close parentheses space colon space open parentheses straight x squared minus 2 straight x minus 15 close parentheses space sisa space open parentheses 4 straight x plus 7 close parentheses straight x squared minus 2 straight x minus 15 equals open parentheses straight x minus 5 close parentheses open parentheses straight x plus 3 close parentheses straight x plus 3 equals 0 rightwards arrow straight x equals negative 3 

Maka:

straight p open parentheses straight x close parentheses equals Pembagi times Hasil space bagi plus sisa straight p open parentheses straight x close parentheses equals Pembagi times straight H open parentheses straight x close parentheses plus straight S open parentheses straight x close parentheses straight p open parentheses straight x close parentheses equals open parentheses straight x squared minus 2 straight x minus 15 close parentheses times straight H open parentheses straight x close parentheses plus open parentheses 4 straight x plus 7 close parentheses straight p open parentheses straight x close parentheses equals open parentheses straight x minus 5 close parentheses open parentheses straight x plus 3 close parentheses times straight H open parentheses straight x close parentheses plus open parentheses 4 straight x plus 7 close parentheses 

polinomial straight p open parentheses straight x close parentheses dibagi oleh open parentheses x plus 3 close parentheses, sehingga:

space space space space straight p open parentheses straight x close parentheses equals Pembagi times Hasil space bagi plus sisa space space space space straight p open parentheses straight x close parentheses equals Pembagi times straight H open parentheses straight x close parentheses plus straight S open parentheses straight x close parentheses space space space space straight p open parentheses straight x close parentheses equals open parentheses straight x squared minus 2 straight x minus 15 close parentheses times straight H open parentheses straight x close parentheses plus open parentheses 4 straight x plus 7 close parentheses space space space space straight p open parentheses straight x close parentheses equals open parentheses straight x minus 5 close parentheses open parentheses straight x plus 3 close parentheses times straight H open parentheses straight x close parentheses plus open parentheses 4 straight x plus 7 close parentheses straight p open parentheses negative 3 close parentheses equals open parentheses negative 3 minus 5 close parentheses open parentheses negative 3 plus 3 close parentheses times straight H open parentheses straight x close parentheses plus open parentheses 4 times negative 3 plus 7 close parentheses straight p open parentheses negative 3 close parentheses equals open parentheses negative 8 close parentheses open parentheses 0 close parentheses times straight H open parentheses straight x close parentheses plus open parentheses negative 12 plus 7 close parentheses straight p open parentheses negative 3 close parentheses equals 0 plus open parentheses negative 5 close parentheses straight p open parentheses negative 3 close parentheses equals 0 minus 5 straight p open parentheses negative 3 close parentheses equals negative 5 

Jadi, jawaban yang benar adalah D.

Roboguru

Sebuah polinomial  berderajat . Polinomial dibagi  bersisa  dan dibagi  bersisa . Polinomial  adalah ....

Pembahasan Soal:

Perhatikan pemfaktoran berikut ini!

begin mathsize 14px style x squared minus 3 x plus 2 equals left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis end style

begin mathsize 14px style x squared minus x minus 6 equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis end style

Diketahui bahwa P(x) dibagi begin mathsize 14px style x squared minus 3 x plus 2 end style sisa begin mathsize 14px style left parenthesis 4 x minus 6 right parenthesis end style, artinya

begin mathsize 14px style p left parenthesis 1 right parenthesis equals 4 left parenthesis 1 right parenthesis minus 6 equals negative 2 p left parenthesis 2 right parenthesis equals 4 left parenthesis 2 right parenthesis minus 6 equals 2 end style

Kemudian, P(x) dibagi begin mathsize 14px style x squared minus x minus 6 end style sisa begin mathsize 14px style left parenthesis 8 x minus 10 right parenthesis end style, artinya

begin mathsize 14px style p left parenthesis 3 right parenthesis equals 8 left parenthesis 3 right parenthesis minus 10 equals 14 p left parenthesis negative 2 right parenthesis equals 8 left parenthesis negative 2 right parenthesis minus 10 equals negative 26 end style

Misalkan begin mathsize 14px style p left parenthesis x right parenthesis equals a x cubed plus b x squared plus c x plus d end style, maka didapat persamaan berikut ini.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell p left parenthesis 1 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row cell a plus b plus c plus d end cell equals cell negative 2 space space space space space space space space space space space space space space space horizontal ellipsis space space left parenthesis 1 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell p left parenthesis 2 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row cell 8 a plus 4 b plus 2 c plus d end cell equals cell 2 space space space space space space space space space space space space space horizontal ellipsis space left parenthesis 2 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell p left parenthesis 3 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row cell 27 a plus 9 b plus 3 c plus d end cell equals cell 14 space space space space space space space space space space horizontal ellipsis space left parenthesis 3 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell p left parenthesis negative 2 right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row cell negative 8 a plus 4 b minus 2 c plus d end cell equals cell negative 26 space space horizontal ellipsis space left parenthesis 4 right parenthesis end cell end table end style  

Eliminasi d

Kurangkan masing-masing persamaan (2), (3), (4) dengan persamaan (1) sehingga diperoleh persamaan baru berikut ini.

Persamaan (2) dikurangi dengan persamaan (1), maka didapat persamaan berikut ini.

begin mathsize 14px style 7 a plus 3 b plus c equals 4 space space... space space open parentheses 5 close parentheses end style 

Persamaan (3) dikurangi dengan persamaan (1), maka didapat persamaan berikut ini.

begin mathsize 14px style 26 a plus 8 b plus 2 c equals 16 space space... space left parenthesis 6 right parenthesis end style 

Persamaan (4) dikurangi dengan persamaan (1), maka didapat persamaan berikut ini.

size 14px minus size 14px 9 size 14px a size 14px plus size 14px 3 size 14px b size 14px minus size 14px 3 size 14px c size 14px equals size 14px minus size 14px 24 size 14px space size 14px space size 14px. size 14px. size 14px. size 14px space size 14px left parenthesis size 14px 7 size 14px right parenthesis 

Eliminasi c

begin mathsize 12px style stack attributes charalign center stackalign right end attributes row 26 a plus 8 b plus 2 c equals 16 open vertical bar cross times 1 close vertical bar 26 a plus 8 b plus 2 c equals 16 none end row row 7 a plus 3 b plus c equals 4 none open vertical bar cross times 2 close vertical bar 14 a plus 6 b plus 2 c equals 8 none minus end row horizontal line row 12 a plus 2 b equals 8 none none end row end stack end style

 

begin mathsize 12px style stack attributes charalign center stackalign right end attributes row minus 9 a plus 3 b minus 3 c equals negative 24 vertical line cross times 1 vertical line minus 9 a plus 3 b minus 3 c equals negative 24 none end row row 7 a plus 3 b plus c equals 4 none vertical line cross times 3 vertical line 21 a plus 9 b plus 3 c equals 12 none plus end row horizontal line row 12 a plus 12 b equals negative 12 none end row end stack end style 

Oleh karena itu, didapat 

begin mathsize 14px style 12 a plus 2 b equals 8 space space space space space space space space space... space space open parentheses 8 close parentheses 12 a plus 12 b equals negative 12 space space... space space open parentheses 9 close parentheses end style 

Kemudian, dari persamaan (8) dan persamaan (9) didapat nilai b berikut ini.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 10 b end cell equals cell negative 20 end cell row b equals cell negative 2 end cell end table end style

Substitusi begin mathsize 14px style b equals negative 2 space end styleke persamaan (8) hingga diperoleh nilai a berikut ini.

begin mathsize 14px style 12 a plus 2 left parenthesis negative 2 right parenthesis equals 8 left right double arrow 12 a equals 12 left right double arrow a equals 1 end style

Substitusi begin mathsize 14px style a equals 1 end style dan begin mathsize 14px style b equals negative 2 space end style ke persamaan (5) diperoleh nilai c berikut ini.

begin mathsize 14px style 7 left parenthesis 1 right parenthesis plus 3 left parenthesis negative 2 right parenthesis plus c equals 4 left right double arrow 1 plus c equals 4 left right double arrow c equals 3 end style

Substitusi a, b, dan c ke (1) diperoleh nilai d berikut ini.

begin mathsize 14px style 1 plus left parenthesis negative 2 right parenthesis plus 3 plus d equals negative 2 left right double arrow 2 plus d equals negative 2 left right double arrow d equals negative 4 end style

Diperoleh begin mathsize 14px style p left parenthesis x right parenthesis equals x cubed minus 2 x squared plus 4 x minus 4 end style.

Jadi, jawaban yang paling tepat adalah A.

Roboguru

Polinomial  dibagi  menghasilkan sisa . Jika  dibagi , menghasilkan sisa . Jika  dibagi , sisanya . Sisa pembagian  oleh  adalah ....

Pembahasan Soal:

Ingat : Sisa pembagian polinomial undefined oleh begin mathsize 14px style open parentheses x minus k close parentheses end style adalah begin mathsize 14px style s equals p left parenthesis k right parenthesis end style   

  • Sisa pembagian polinomial undefined oleh begin mathsize 14px style open parentheses x minus 1 close parentheses end style adalah begin mathsize 14px style 24 end style atau dapat ditulis  begin mathsize 14px style p left parenthesis 1 right parenthesis equals 24 end style 
  • Sisa pembagian polinomial undefined oleh begin mathsize 14px style open parentheses x plus 1 close parentheses end style adalah begin mathsize 14px style 8 end style atau dapat ditulis begin mathsize 14px style p left parenthesis negative 1 right parenthesis equals 8 end style   
  • Sisa pembagian polinomial begin mathsize 14px style p left parenthesis x right parenthesis end style oleh begin mathsize 14px style open parentheses x minus 3 close parentheses end style adalah begin mathsize 14px style 32 end style atau dapat ditulis begin mathsize 14px style p left parenthesis 3 right parenthesis equals 32 end style 
  • Sisa pembagian polinomial undefined oleh begin mathsize 14px style open parentheses x squared minus 1 close parentheses open parentheses x minus 3 close parentheses end style adalah begin mathsize 14px style a x squared plus b x plus c end style 
    karena pembaginya berderajat 3, maka sisa pembagiannya berderajat ≤ (3 – 1) ⇒ berderajat ≤ 2

begin mathsize 14px style open parentheses x squared minus 1 close parentheses open parentheses x minus 3 close parentheses equals open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses open parentheses x minus 3 close parentheses end style

Untuk begin mathsize 14px style x equals 1 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p left parenthesis x right parenthesis end cell equals cell a x squared plus b x plus c end cell row cell p left parenthesis 1 right parenthesis end cell equals 24 row cell a times 1 squared plus b times 1 plus c end cell equals 24 row cell a plus b plus c end cell equals cell 24 space horizontal ellipsis space pers space left parenthesis 1 right parenthesis end cell end table end style 

Untuk begin mathsize 14px style x equals negative 1 end style 

table attributes columnalign right center left columnspacing 0px end attributes row cell size 14px p size 14px left parenthesis size 14px x size 14px right parenthesis end cell size 14px equals cell size 14px a size 14px x to the power of size 14px 2 size 14px plus size 14px b size 14px x size 14px plus size 14px c end cell row cell size 14px p size 14px left parenthesis size 14px minus size 14px 1 size 14px right parenthesis end cell size 14px equals size 14px 8 row cell size 14px a size 14px times begin mathsize 14px style left parenthesis negative 1 right parenthesis end style to the power of size 14px 2 size 14px plus size 14px b size 14px times begin mathsize 14px style left parenthesis negative 1 right parenthesis end style size 14px plus size 14px c end cell size 14px equals size 14px 8 row cell size 14px a size 14px minus size 14px b size 14px plus size 14px c end cell size 14px equals cell size 14px 8 size 14px space size 14px horizontal ellipsis size 14px space size 14px pers size 14px space size 14px left parenthesis size 14px 2 size 14px right parenthesis end cell end table 

Untuk begin mathsize 14px style x equals 3 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p left parenthesis x right parenthesis end cell equals cell a x squared plus b x plus c end cell row cell p left parenthesis 3 right parenthesis end cell equals 32 row cell a times left parenthesis 3 right parenthesis squared plus b times left parenthesis 3 right parenthesis plus c end cell equals 32 row cell 9 a plus 3 b plus c end cell equals cell 32 space horizontal ellipsis space pers space left parenthesis 3 right parenthesis end cell end table end style 

Eliminasi a dan c persamaan (1) dan (2) 

begin mathsize 14px style a plus b plus c equals 24 bottom enclose a minus b plus c equals 8 end enclose space minus space space space space space space space space space 2 b equals 16 space space space space space space space space space space space b equals 8 end style 

Substitusikan b = 8 ke persamaan (1)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a plus b plus c end cell equals 24 row cell a plus 8 plus c end cell equals 24 row c equals cell 24 minus 8 minus a end cell row c equals cell 16 minus a end cell end table end style 

Substitusikan b = 8 dan c = 16 – a ke persamaan (3)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 a plus 3 b plus c end cell equals 32 row cell 9 a plus 3 times 8 plus 16 minus a end cell equals 32 row cell 8 a end cell equals cell 32 minus 24 minus 16 end cell row cell 8 a end cell equals cell negative 8 end cell row a equals cell negative 1 end cell end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row c equals cell 16 minus a end cell row blank equals cell 16 minus left parenthesis negative 1 right parenthesis end cell row blank equals 17 end table end style 

Maka, Sisa pembagian undefined oleh undefined  adalahbegin mathsize 14px style negative x squared plus 8 x plus 17 end style   

Jadi, Jawaban yang tepat adalah C.

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