Soal dan Pembahasan Buku Sukino BAB 4 Polinomial LKS 5 Matematika Peminatan Kelas XI Kurikulum 2013

LKS 5

1.   Tuliskan empat suku pertama dalam pangkat naik dari x pada setiap penjabaran (ekspansi) berikut:

a.  \({\left( {1 + \frac{x}{2}} \right)^6}\)

b.  \({\left( {1 + 2{x^2}} \right)^9}\)

c.   \({\left( {x + \frac{1}{x}} \right)^8}\)

d.   \({\left( {1 + x} \right)^2}{\left( {2 + x} \right)^6}\)

e.  \(x(1 + 3x){\left( {1 + \frac{1}{x}} \right)^6}\)

f.   \({\left( {1 + 3x} \right)^2}{\left( {2 + x} \right)^6}\)

Pembahahasan:
a.   \({\left( {1 + \frac{x}{2}} \right)^6} = C_0^6.{\left( 1 \right)^6}.{\left( {\frac{x}{2}} \right)^0} + C_1^6.{\left( 1 \right)^5}.{\left( {\frac{x}{2}} \right)^1} + C_2^6.{\left( 1 \right)^4}.{\left( {\frac{x}{2}} \right)^2} + C_3^6.{\left( 1 \right)^3}.{\left( {\frac{x}{2}} \right)^3} +  \ldots \)
\( = 1.1.1 + 6.1.\frac{x}{2} + 15.1.\frac{{{x^2}}}{4} + 20.1.\frac{{{x^3}}}{8} +  \ldots \)
\( = 1 + 3x + \frac{{15}}{4}{x^2} + \frac{5}{2}{x^3} +  \ldots \)

b.  \({\left( {1 + 2{x^2}} \right)^9} = C_0^9.{\left( 1 \right)^9}.{\left( {2{x^2}} \right)^0} + C_1^9.{\left( 1 \right)^8}.{\left( {2{x^2}} \right)^1} + C_2^9.{\left( 1 \right)^7}.{\left( {2{x^2}} \right)^2} + C_3^9.{\left( 1 \right)^6}.{\left( {2{x^2}} \right)^3} +  \ldots \)
\( = 1.1.1 + 9.1.2{x^2} + 36.1.4{x^4} + 84.1.8{x^6} +  \ldots \)
\( = 1 + 18{x^2} + 144{x^4} + 672{x^6} +  \ldots \)

c.  \({\left( {x + \frac{1}{x}} \right)^8} = C_8^8.{\left( x \right)^0}.{\left( {\frac{1}{x}} \right)^8} + C_7^8.{\left( x \right)^1}.{\left( {\frac{1}{x}} \right)^7} + C_6^8.{\left( x \right)^2}.{\left( {\frac{1}{x}} \right)^6} + C_5^8.{\left( x \right)^3}.{\left( {\frac{1}{x}} \right)^5} +  \ldots \)
\( = 1.1.\frac{1}{{{x^8}}} + 8.x.\frac{1}{{{x^7}}} + 28.{x^2}.\frac{1}{{{x^6}}} + 56.{x^3}.\frac{1}{{{x^5}}} +  \ldots \)
\( = {x^{ - 8}} + 8{x^{ - 6}} + 28{x^{ - 4}} + 56{x^{ - 2}} +  \ldots \)

d.  \({\left( {1 + x} \right)^2}{\left( {2 + x} \right)^6} = \left( {1 + 2x + {x^2}} \right)(C_0^6.{\left( 2 \right)^6}.{\left( x \right)^0} + C_1^6.{\left( 2 \right)^5}.{\left( x \right)^1} + C_2^6.{\left( 2 \right)^4}.{\left( x \right)^2} + C_3^6.{\left( 2 \right)^3}.{\left( x \right)^3} + C_4^6.{\left( 2 \right)^2}.{\left( x \right)^4} + C_5^6.{\left( 2 \right)^1}.{\left( x \right)^5} + C_6^6.{\left( 2 \right)^0}.{\left( x \right)^6})\)
\( = \left( {1 + 2x + {x^2}} \right)\left( {1.64.1 + 6.32.x + 15.16.{x^2} + 20.8.{x^3} + 15.4.{x^4} + 6.2.{x^5} + 1.1.{x^6}} \right)\)
\( = \left( {1 + 2x + {x^2}} \right)\left( {64 + 192x + 240{x^2} + 160{x^3} + 60{x^4} + 12{x^5} + {x^6}} \right)\)
Variabel dari empat suku pertama dalam pangkat naik pada operasi polynomial ini adalah \({x^0},{x^1},{x^2},{x^3}\) yaitu:
\( = 64 + 192x + 128x + 384{x^2} + 240{x^2} + 64{x^2} + 160{x^3} + 480{x^3} + 192{x^3} +  \ldots \)
\( = 64 + 320x + 688{x^2} + 832{x^3} +  \ldots \)

e.  \(x\left( {1 + 3x} \right){\left( {1 + \frac{1}{x}} \right)^6} = \left( {x + 3{x^2}} \right)\left( {C_6^6.{{\left( 1 \right)}^0}.{{\left( {\frac{1}{x}} \right)}^6} + C_5^6.{{\left( 1 \right)}^1}.{{\left( {\frac{1}{x}} \right)}^5} + C_4^6.{{\left( 1 \right)}^2}.{{\left( {\frac{1}{x}} \right)}^4} + C_3^6.{{\left( 1 \right)}^3}.{{\left( {\frac{1}{x}} \right)}^3} + } \right.\left. {C_2^6.{{\left( 1 \right)}^4}.{{\left( {\frac{1}{x}} \right)}^2} + C_1^6.{{\left( 1 \right)}^5}.{{\left( {\frac{1}{x}} \right)}^1} + C_0^6.{{\left( 1 \right)}^6}.{{\left( {\frac{1}{x}} \right)}^0}} \right)\)
\( = \left( {x + 3{x^2}} \right)\left( {1.1.\frac{1}{{{x^6}}} + 6.1.\frac{1}{{{x^5}}} + 15.1.\frac{1}{{{x^4}}} + 20.1.\frac{1}{{{x^3}}} + } \right.\left. {15.1.\frac{1}{{{x^2}}} + 6.1.\frac{1}{x} + 1.1.1} \right)\)
\( = \left( {x + 3{x^2}} \right)\left( {{x^{ - 6}} + 6{x^{ - 5}} + 15{x^{ - 4}} + 20{x^{ - 3}} + } \right.\left. {15{x^{ - 2}} + 6{x^{ - 1}} + 1} \right)\)
Variabel dari empat suku pertama dalam pangkat naik pada operasi polynomial ini adalah \({x^{ - 5}},{x^{ - 4}},{x^{ - 3}},{x^{ - 2}}\) yaitu:
\( = {x^{ - 5}} + 6{x^{ - 4}} + 3{x^{ - 4}} + 15{x^{ - 3}} + 18{x^{ - 3}} + 20{x^{ - 2}} + 45{x^{ - 2}} +  \ldots \)
\( = {x^{ - 5}} + 9{x^{ - 4}} + 33{x^{ - 3}} + 65{x^{ - 2}} +  \ldots \)

f.  \({\left( {1 + 3x} \right)^2}{\left( {2 + x} \right)^6} = \left( {1 + 6x + 9{x^2}} \right)(C_0^6.{\left( 2 \right)^6}.{\left( x \right)^0} + C_1^6.{\left( 2 \right)^5}.{\left( x \right)^1} + C_2^6.{\left( 2 \right)^4}.{\left( x \right)^2} + C_3^6.{\left( 2 \right)^3}.{\left( x \right)^3} + C_4^6.{\left( 2 \right)^2}.{\left( x \right)^4} + C_5^6.{\left( 2 \right)^1}.{\left( x \right)^5} + C_6^6.{\left( 2 \right)^0}.{\left( x \right)^6})\)
\( = \left( {1 + 6x + 9{x^2}} \right)\left( {1.64.1 + 6.32.x + 15.16.{x^2} + 20.8.{x^3} + 15.4.{x^4} + 6.2.{x^5} + 1.1.{x^6}} \right)\)
\( = \left( {1 + 6x + 9{x^2}} \right)\left( {64 + 192x + 240{x^2} + 160{x^3} + 60{x^4} + 12{x^5} + {x^6}} \right)\)
Variabel dari empat suku pertama dalam pangkat naik pada operasi polynomial ini adalah \({x^0},{x^1},{x^2},{x^3}\) yaitu:
\( = 64 + 192x + 384x + 1152{x^2} + 240{x^2} + 576{x^2} + 160{x^3} + 1440{x^3} + 1728{x^3} +  \ldots \)
\( = 64 + 576x + 1968{x^2} + 3328{x^3} +  \ldots \)

2.   Tentukan koefisien dari \({x^4}\) pada setiap penjabaran berikut:

a.  \({\left( {1 + \frac{x}{2}} \right)^{10}}\)

b.   \({\left( {x + \frac{1}{x}} \right)^9}\)

c.    \({\left( {1 + 2{x^2}} \right)^{11}}\)

d.   \({\left( {1 + 3x} \right)^3}{\left( {2 + x} \right)^6}\)

Pembahasan :
a.  \({\left( {1 + \frac{x}{2}} \right)^{10}} =  \ldots  + C_4^{10}.{\left( 1 \right)^6}.{\left( {\frac{x}{2}} \right)^4} +  \ldots  =  \ldots  + 210.1.\frac{{{x^4}}}{{16}} +  \ldots  =  \ldots  + \frac{{105}}{8}{x^4} +  \ldots \)
Jadi, koefisien \({x^4}\) dari penjabaran \({\left( {1 + \frac{x}{2}} \right)^{10}}\) adalah \(\frac{{105}}{8}\)

b.  \({\left( {x + \frac{1}{x}} \right)^9} = C_0^9.{\left( x \right)^9}.{\left( {\frac{1}{x}} \right)^0} + C_1^9.{\left( x \right)^8}.{\left( {\frac{1}{x}} \right)^1} + C_2^9.{\left( x \right)^7}.{\left( {\frac{1}{x}} \right)^2} + C_3^9.{\left( x \right)^6}.{\left( {\frac{1}{x}} \right)^3} + C_4^9.{\left( x \right)^5}.{\left( {\frac{1}{x}} \right)^4} + C_5^9.{\left( x \right)^4}.{\left( {\frac{1}{x}} \right)^5} + C_6^9.{\left( x \right)^3}.{\left( {\frac{1}{x}} \right)^6} + C_7^9.{\left( x \right)^2}.{\left( {\frac{1}{x}} \right)^7} + C_8^9.{\left( x \right)^1}.{\left( {\frac{1}{x}} \right)^8} + C_9^9.{\left( x \right)^0}.{\left( {\frac{1}{x}} \right)^9}\)
\( = 1.{x^9}.1 + 9.{x^8}.\frac{1}{x} + 36.{x^7}.\frac{1}{{{x^2}}} + 84.{x^6}.\frac{1}{{{x^3}}} + 126.{x^5}.\frac{1}{{{x^4}}} + 126.{x^4}.\frac{1}{{{x^5}}} + 84.{x^3}.\frac{1}{{{x^6}}} + 36.{x^2}.\frac{1}{{{x^7}}} + 9.{x^1}.\frac{1}{{{x^8}}} + 1.1.\frac{1}{{{x^9}}}\)
\( = {x^9} + 9{x^7} + 36{x^5} + 84{x^3} + 126x + 126{x^{ - 1}} + 84{x^{ - 3}} + 36{x^{ - 5}} + 9{x^{ - 7}} + {x^{ - 9}}\)
Dari penjabaran, dapat dilihat bahwa tidak ada variabel \({x^4}\). Artinya koefisien\({x^4}\) dari penjabaran \({\left( {x + \frac{1}{x}} \right)^9}\) adalah 0

c.  \({\left( {1 + 2{x^2}} \right)^{11}} =  \ldots  + C_2^{11}.{\left( 1 \right)^9}.{\left( {2{x^2}} \right)^2} +  \ldots  =  \ldots  + 55.1.4{x^4} +  \ldots  =  \ldots  + 220{x^4} +  \ldots \)
Jadi, koefisien \({x^4}\) dari penjabaran \({\left( {1 + 2{x^2}} \right)^{11}}\) adalah 220

d.  \[{\left( {1 + 3x} \right)^3}{\left( {2 + x} \right)^6} = \left( {C_0^3.{{\left( 1 \right)}^3}.{{\left( {3x} \right)}^0}} \right. + C_1^3.{\left( 1 \right)^2}.{\left( {3x} \right)^1} + C_2^3.{\left( 1 \right)^1}.{\left( {3x} \right)^2} + C_3^3.\left. {{{\left( 1 \right)}^0}.{{\left( {3x} \right)}^3}} \right)\left( {C_0^6.{{\left( 2 \right)}^6}.{{\left( x \right)}^0} + } \right.C_1^6.{\left( 2 \right)^5}.{\left( x \right)^1} + C_2^6.{\left( 2 \right)^4}.{\left( x \right)^2} + C_3^6.{\left( 2 \right)^3}.{\left( x \right)^3} + C_4^6.{\left( 2 \right)^2}.{\left( x \right)^4} + C_5^6.{\left( 2 \right)^1}.{\left( x \right)^5} + C_6^6.{\left( 2 \right)^0}.{\left( x \right)^6})\]
\( = \left( {1 + 9x + 9{x^2} + 27{x^3}} \right)\left( {1.64.1 + 6.32.x + 15.16.{x^2} + 20.8.{x^3} + 15.4.{x^4} + 6.2.{x^5} + 1.1.{x^6}} \right)\)
\( = \left( {1 + 9x + 9{x^2} + 27{x^3}} \right)\left( {64 + 192x + 240{x^2} + 160{x^3} + 60{x^4} + 12{x^5} + {x^6}} \right)\)
\( =  \ldots  + 60{x^4} + 1440{x^4} + 2160{x^4} + 5184{x^4} +  \ldots \)
\( =  \ldots  + 8844{x^4} +  \ldots \)
Jadi, koefisien \({x^4}\) dari penjabaran \({\left( {1 + 2{x^2}} \right)^{11}} adalah 8844




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