# Akar Dari 117

Akar Dari 117

Human relationship between the rational roots of a polynomial and its extreme coefficients

In algebra, the
rational root theorem

(or
rational root exam,
rational nil theorem,
rational zero examination

or

p/q

theorem
) states a constraint on rational solutions of a polynomial equation

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

a

due due north

10

north

+

a

due north

1

x

due northward

one

+

+

a

=

{\displaystyle a_{north}10^{north}+a_{north-1}ten^{northward-1}+\cdots +a_{0}=0}

with integer coefficients

${\displaystyle a_{i}\in \mathbb {Z} }$

a

i

Z

{\displaystyle a_{i}\in \mathbb {Z} }

and

${\displaystyle a_{0},a_{n}\neq 0}$

a

,

a

n

{\displaystyle a_{0},a_{n}\neq 0}

. Solutions of the equation are too chosen roots or zeroes of the polynomial on the left side.

The theorem states that each rational solution

10 =

pq

, written in everyman terms and so that

p

and

q

are relatively prime, satisfies:

• p

is an integer cistron of the abiding term

a

, and

• q

is an integer factor of the leading coefficient

.

The rational root theorem is a special case (for a unmarried linear gene) of Gauss’s lemma on the factorization of polynomials. The
integral root theorem

is the special example of the rational root theorem when the leading coefficient is

= ane
.

## Application

The theorem is used to notice all rational roots of a polynomial, if any. Information technology gives a finite number of possible fractions which tin can exist checked to see if they are roots. If a rational root

x

=
r

is found, a linear polynomial
(10

r)

tin can be factored out of the polynomial using polynomial long sectionalisation, resulting in a polynomial of lower caste whose roots are besides roots of the original polynomial.

### Cubic equation

The full full general cubic equation

${\displaystyle ax^{3}+bx^{2}+cx+d=0}$

a

x

iii

+

b

x

ii

+

c

ten

+

d

=

{\displaystyle ax^{iii}+bx^{ii}+cx+d=0}

with integer coefficients has three solutions in the circuitous plane. If the rational root test finds no rational solutions, then the but style to limited the solutions algebraically uses cube roots. Only if the exam finds a rational solution

r

, and and so factoring out
(ten

r)

leaves a quadratic polynomial whose two roots, constitute with the quadratic formula, are the remaining ii roots of the cubic, fugitive cube roots.

## Proofs

### Simple proof

Allow

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

P

(

10

)

=

a

n

x

n

+

a

due north

one

10

n

1

+

+

a

one

x

+

a

{\displaystyle P(ten)\ =\ a_{n}ten^{northward}+a_{due north-ane}x^{north-ane}+\cdots +a_{1}x+a_{0}}

with

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

a

,

a

due north

Z

.

{\displaystyle a_{0},\ldots a_{n}\in \mathbb {Z} .}

Suppose

P(p/q) = 0

for some coprime

p,
q

:

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

P

(

p

q

)

=

a

northward

(

p

q

)

north

+

a

n

1

(

p

q

)

n

1

+

+

a

one

(

p

q

)

+

a

=

0.

{\displaystyle P\left({\tfrac {p}{q}}\right)=a_{north}\left({\tfrac {p}{q}}\correct)^{north}+a_{north-1}\left({\tfrac {p}{q}}\right)^{n-i}+\cdots +a_{i}\left({\tfrac {p}{q}}\correct)+a_{0}=0.}

To clear denominators, multiply both sides by

q

north

:

${\displaystyle a_{n}p^{n}+a_{n-1}p^{n-1}q+\cdots +a_{1}pq^{n-1}+a_{0}q^{n}=0.}$

a

northward

p

north

+

a

due north

one

p

n

1

q

+

+

a

1

p

q

northward

1

+

a

q

due north

=

0.

{\displaystyle a_{n}p^{n}+a_{northward-ane}p^{n-1}q+\cdots +a_{one}pq^{n-one}+a_{0}q^{north}=0.}

Shifting the

a

term to the right side and factoring out
p

on the left side produces:

${\displaystyle p\left(a_{n}p^{n-1}+a_{n-1}qp^{n-2}+\cdots +a_{1}q^{n-1}\right)=-a_{0}q^{n}.}$

p

(

a

north

p

due north

1

+

a

n

ane

q

p

due north

2

+

+

a

one

q

due north

i

)

=

a

q

n

.

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

Thus,
p

divides

a

qdue north

. But
p

is coprime to
q

and therefore to

qnorthward

, so past Euclid’s lemma
p

must divide the remaining cistron

a

.

On the other hand, shifting the

a

northward

term to the right side and factoring out
q

on the left side produces:

${\displaystyle q\left(a_{n-1}p^{n-1}+a_{n-2}qp^{n-2}+\cdots +a_{0}q^{n-1}\right)=-a_{n}p^{n}.}$

q

(

a

due northward

1

p

n

ane

+

a

due north

2

q

p

n

two

+

+

a

q

northward

1

)

=

a

n

p

due north

.

{\displaystyle q\left(a_{n-ane}p^{due north-one}+a_{n-2}qp^{due north-ii}+\cdots +a_{0}q^{n-1}\right)=-a_{north}p^{n}.}

Reasoning as before, it follows that
q

divides

.[one]

### Proof using Gauss’ lemma

Should there exist a nontrivial gene dividing all the coefficients of the polynomial, then one can separate by the greatest common divisor of the coefficients so as to obtain a primitive polynomial in the sense of Gauss’south lemma; this does non alter the set upwardly of rational roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in

Q[X]
, then it besides factors in

Z[10]

as a product of archaic polynomials. At present any rational root

p/q

corresponds to a gene of degree 1 in

Q[Ten]

of the polynomial, and its primitive representative is then

qx

p

, bold that

p

and

q

are coprime. Merely whatever multiple in

Z[ten]

of

qx

p

q

and abiding term divisible by

p

, which proves the statement. This argument shows that more than than generally, any irreducible factor of

P

can be supposed to take integer coefficients, and leading and abiding coefficients dividing the corresponding coefficients of

P

.

## Examples

### First

In the polynomial

${\displaystyle 2x^{3}+x-1,}$

2

10

3

+

x

i

,

{\displaystyle 2x^{3}+x-ane,}

whatever rational root fully reduced would take to have a numerator that divides evenly into one and a denominator that divides evenly into 2. Hence the only possible rational roots are ±i/two and ±1; since neither of these equates the polynomial to zilch, it has no rational roots.

### Second

In the polynomial

${\displaystyle x^{3}-7x+6}$

x

3

vii

ten

+

six

{\displaystyle x^{three}-7x+half dozen}

the but possible rational roots would take a numerator that divides 6 and a denominator that divides one, limiting the possibilities to ±ane, ±two, ±three, and ±6. Of these, 1, 2, and –3 equate the polynomial to zero, and hence are its rational roots. (In fact these are its but roots since a cubic has but 3 roots; in general, a polynomial could have some rational and some irrational roots.)

### third

Every rational root of the polynomial

${\displaystyle 3x^{3}-5x^{2}+5x-2}$

three

ten

iii

v

x

2

+

5

x

ii

{\displaystyle 3x^{three}-5x^{2}+5x-2}

must be amidst the numbers symbolically indicated by:

${\displaystyle \pm {\tfrac {1,2}{1,3}}=\pm \left\{1,2,{\tfrac {1}{3}},{\tfrac {2}{3}}\right\}.}$

±

one

,

2

1

,

3

=

±

{

i

,

2

,

i

three

,

ii

3

}

.

{\displaystyle \pm {\tfrac {ane,ii}{one,3}}=\pm \left\{i,ii,{\tfrac {i}{3}},{\tfrac {2}{three}}\correct\}.}

These 8 root candidates

10

=
r

can be tested by evaluating

P(r)
, for case using Horner’s method. It turns out at that place is exactly one with

P(r) = 0
.

This procedure may be fabricated more than efficient: if

P(r) ≠ 0
, information technology can be used to shorten the list of remaining candidates.[ii]

For case,

x

= 1

does not piece of work, as

P(ane) = i
. Substituting

10

= ane +
t

yields a polynomial int

with abiding term

P(i) = one
, while the coefficient of

t

3

remains the aforementioned every bit the coefficient of

x

3

. Applying the rational root theorem thus yields the possible roots

${\displaystyle t=\pm {\tfrac {1}{1,3}}}$

t

=

±

one

1

,

three

{\displaystyle t=\pm {\tfrac {one}{one,3}}}

, and then that

${\displaystyle x=1+t=2,0,{\tfrac {4}{3}},{\tfrac {2}{3}}.}$

10

=

1

+

t

=

two

,

,

four

iii

,

two

3

.

{\displaystyle x=1+t=ii,0,{\tfrac {four}{3}},{\tfrac {two}{three}}.}

Truthful roots must occur on both lists, so listing of rational root candidates has shrunk to only

ten

= 2

and

x

= ii/iii
.

If

1000

≥ one

rational roots are found, Horner’southward method volition too yield a polynomial of caste

due north

k

whose roots, together with the rational roots, are exactly the roots of the original polynomial. If none of the candidates is a solution, at that place tin can be no rational solution.

## Run across also

• Fundamental theorem of algebra
• Integrally airtight domain
• Descartes’ dominion of signs
• Gauss–Lucas theorem
• Backdrop of polynomial roots
• Content (algebra)
• Eisenstein’southward criterion

## Notes

1. ^

Arnold, D.; Arnold, G. (1993).
Iv unit mathematics. Edward Arnold. pp. 120–121. ISBN0-340-54335-iii.

2. ^

King, Jeremy D. (Nov 2006). “Integer roots of polynomials”.
Mathematical Gazette.
xc: 455–456.

## References

• Charles D. Miller, Margaret L. Lial, David I. Schneider:
Fundamentals of Higher Algebra. Scott & Foresman/Trivial & Dark-brown Higher Instruction, 3rd edition 1990, ISBN 0-673-38638-4, pp. 216–221
• Phillip Due south. Jones, Jack D. Bedient:
The historical roots of uncomplicated mathematics. Dover Courier Publications 1998, ISBN 0-486-25563-eight, pp. 116–117 (online re-create, p. 116, at Google Books)
• Ron Larson:
Calculus: An Practical Approach. Cengage Learning 2007, ISBN 978-0-618-95825-2, pp. 23–24 (online copy, p. 23, at Google Books)

### Akar Dari 117

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