# Akar Rasional

Akar Rasional.

Dari Wikipedia bahasa Indonesia, ensiklopedia bebas

${\displaystyle x^{3}-4x^{2}+2x-8=0}$

x

3

4

10

ii

+
2
ten

viii
=

{\displaystyle 10^{iii}-4x^{two}+2x-eight=0}

Nilai

${\displaystyle x}$

ten

{\displaystyle x}

Nilai

${\displaystyle P(x)}$

P
(
x
)

{\displaystyle P(x)}

${\displaystyle -8}$

eight

{\displaystyle -8}

${\displaystyle -792}$

792

{\displaystyle -792}

${\displaystyle -4}$

iv

{\displaystyle -4}

${\displaystyle -144}$

144

{\displaystyle -144}

${\displaystyle -2}$

2

{\displaystyle -2}

${\displaystyle -36}$

36

{\displaystyle -36}

${\displaystyle -1}$

ane

{\displaystyle -one}

${\displaystyle -15}$

xv

{\displaystyle -15}

${\displaystyle 1}$

1

{\displaystyle ane}

${\displaystyle -9}$

9

{\displaystyle -9}

${\displaystyle 2}$

2

{\displaystyle 2}

${\displaystyle -12}$

12

{\displaystyle -12}

${\displaystyle 4}$

4

{\displaystyle 4}

${\displaystyle 0}$

{\displaystyle 0}

${\displaystyle 8}$

8

{\displaystyle eight}

${\displaystyle 264}$

264

{\displaystyle 264}

Teorema akar rasional
atau
uji akar rasional
[1]
atau
teorema rasional nol
[1]. Teorema ini menjelaskan persamaan polinomial dengan koefisien adalah bilangan bulat dan solusi akarnya berupa bilangan rasional. Teorema mengatakan bahwa untuk persamaan

${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}=0}$

a

due north

ten

n

+

a

northward

one

x

n

1

+

+

a

one

x
+

a

=

{\displaystyle a_{n}ten^{n}+a_{n-1}x^{n-one}+\cdots +a_{one}ten+a_{0}=0}

,

dimana

${\displaystyle a_{0},\dots ,a_{n}\in \mathbb {Z} }$

a

,

,

a

n

Z

{\displaystyle a_{0},\dots ,a_{northward}\in \mathbb {Z} }

. Jika persamaan memiliki suatu akar rasional, maka bentuk akar tersebut adalah

${\displaystyle x=\left\{{\frac {\pm {\text{faktor dari }}a_{0}}{\pm {\text{faktor dari }}a_{n}}}\right\}}$

10
=

{

±

faktor dari

a

±

faktor dari

a

due north

}

{\displaystyle ten=\left\{{\frac {\pm {\text{faktor dari }}a_{0}}{\pm {\text{faktor dari }}a_{n}}}\right\}}

,

asalkan penyebut dan pembilang pada suatu solusi

${\displaystyle x}$

x

{\displaystyle 10}

(adalah bilangan rasional) harus membagi habis

${\displaystyle a_{n}}$

a

n

{\displaystyle a_{n}}

dan

${\displaystyle a_{0}}$

a

{\displaystyle a_{0}}

.

Misalnya, diberikan persamaan

${\displaystyle P(x)=x^{3}-4x^{2}+2x-8=0}$

P
(
10
)
=

x

3

iv

x

2

+
2
x

8
=

{\displaystyle P(ten)=x^{3}-4x^{2}+2x-8=0}

${\displaystyle -8}$

8

{\displaystyle -8}

memiliki faktor

${\displaystyle \pm 1,\pm 2,\pm 4,\pm 8}$

±

ane
,
±

ii
,
±

4
,
±

viii

{\displaystyle \pm 1,\pm 2,\pm four,\pm 8}

dan

${\displaystyle 1}$

1

{\displaystyle ane}

memiliki faktor

${\displaystyle \pm 1}$

±

1

{\displaystyle \pm one}

${\displaystyle \pm \{1,2,4,8\}}$

±

{
1
,
ii
,
4
,
8
}

{\displaystyle \pm \{1,2,4,viii\}}

. Dengan memasukkan semua kemungkinan nilai

${\displaystyle x}$

ten

{\displaystyle ten}

agar persamaan di atas sama dengan nol, maka kita memperoleh

${\displaystyle x=4}$

x
=
4

{\displaystyle x=4}

.

## Bukti

Misal

${\textstyle x={\frac {p}{q}}}$

10
=

p
q

{\textstyle x={\frac {p}{q}}}

${\displaystyle P(x)}$

P
(
x
)

{\displaystyle P(x)}

. Kita cukup membuktikan teorema ini bahwa

${\displaystyle p\mid a_{0}}$

p

a

{\displaystyle p\mid a_{0}}

dan

${\displaystyle q\mid a_{n}}$

q

a

northward

{\displaystyle q\mid a_{due north}}

, dimana

${\displaystyle \operatorname {FPB} (p,q)=1}$

FPB

(
p
,
q
)
=
one

{\displaystyle \operatorname {FPB} (p,q)=ane}

. Substitusi nilai

${\displaystyle x}$

x

{\displaystyle x}

sehingga kita memperoleh

${\displaystyle a_{n}\left({\frac {p}{q}}\right)^{n}+a_{n-1}\left({\frac {p}{q}}\right)^{n-1}+\cdots +a_{1}\left({\frac {p}{q}}\right)+a_{0}=0}$

a

n

(

p
q

)

n

+

a

n

1

(

p
q

)

n

1

+

+

a

1

(

p
q

)

+

a

=

{\displaystyle a_{n}\left({\frac {p}{q}}\right)^{north}+a_{n-one}\left({\frac {p}{q}}\right)^{n-ane}+\cdots +a_{1}\left({\frac {p}{q}}\right)+a_{0}=0}

.

Kita akan membuktikan bahwa

${\displaystyle p}$

p

{\displaystyle p}

membagi habis

${\displaystyle a_{0}}$

a

{\displaystyle a_{0}}

${\displaystyle a_{0}}$

a

{\displaystyle a_{0}}

.

${\displaystyle a_{n}\left({\frac {p}{q}}\right)^{n}+a_{n-1}\left({\frac {p}{q}}\right)^{n-1}+\cdots +a_{1}\left({\frac {p}{q}}\right)=-a_{0}}$

a

due north

(

p
q

)

n

+

a

n

1

(

p
q

)

n

ane

+

+

a

ane

(

p
q

)

=

a

{\displaystyle a_{n}\left({\frac {p}{q}}\right)^{n}+a_{north-i}\left({\frac {p}{q}}\right)^{northward-i}+\cdots +a_{1}\left({\frac {p}{q}}\right)=-a_{0}}

.

Bagi kedua ruas dengan

${\displaystyle q^{n}}$

q

northward

{\displaystyle q^{n}}

dan faktor-keluarkan

${\displaystyle p}$

p

{\displaystyle p}

untuk ruas kiri. Kita memperoleh

${\displaystyle p\left(a_{n}p^{n-1}+\cdots +a_{1}\right)=-a_{0}q^{n}}$

p

(

a

n

p

n

1

+

+

a

one

)

=

a

q

n

{\displaystyle p\left(a_{north}p^{n-1}+\cdots +a_{1}\correct)=-a_{0}q^{due north}}

.

Disini, kita memperoleh bahwa

${\displaystyle p}$

p

{\displaystyle p}

membagi habis

${\displaystyle a_{0}}$

a

{\displaystyle a_{0}}

. Sekarang, kita membuktikan

${\displaystyle q}$

q

{\displaystyle q}

membagi habis

${\displaystyle a_{n}}$

a

n

{\displaystyle a_{n}}

. Dengan cara yang serupa, kita pindah-ruaskan

${\textstyle a_{n}\left({\frac {p}{q}}\right)^{n}}$

a

n

(

p
q

)

n

{\textstyle a_{north}\left({\frac {p}{q}}\right)^{n}}

dan kalikan kedua ruas dengan

${\displaystyle q^{n}}$

q

n

{\displaystyle q^{due north}}

.

${\displaystyle q\left(a_{n-1}p^{n-1}+\cdots +a_{1}pq^{n+1}+a_{0}q^{n}\right)=-a_{n}p^{n}}$

q

(

a

due north

1

p

northward

i

+

+

a

1

p

q

northward
+
ane

+

a

q

northward

)

=

a

n

p

northward

{\displaystyle q\left(a_{n-one}p^{due north-1}+\cdots +a_{one}pq^{n+1}+a_{0}q^{northward}\right)=-a_{due north}p^{n}}

.

Disini, kita memperoleh bahwa

${\displaystyle q}$

q

{\displaystyle q}

membagi habis

${\displaystyle a_{n}}$

a

n

{\displaystyle a_{n}}

.

${\displaystyle \blacksquare }$

{\displaystyle \blacksquare }

[3]

## Rujukan

1. ^

a

b

“Teorema akar rasional | matematika”.
Teorema akar rasional | matematika. 2020-06-27.

2. ^

“Sutori”.
world wide web.sutori.com
(dalam bahasa Inggris). Diakses tanggal
2021-12-23
.

3. ^

“Teorema Akar Rasional”.
ICHI.PRO
. Diakses tanggal
2021-12-twenty
.

## Akar Rasional

Source: https://id.wikipedia.org/wiki/Teorema_akar_rasional

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