Theory Of Equations 1

THEORY OF )EQUATIONS )1

CONTENTS 

  

The function n

f(x) = anx + an−1x

n− 1

+ an− 2 x

n− 2

+

⋅ ⋅ ⋅ + a1x + a0

Is called a polynomial of degree n in x The equation f)x) = 0 is called an algebraic equation in x .

The quantity r such that

f)r) = 0 is called a root of the equation and it may be real or complex

The Remainder Theorem If R is the remainder of dividing f)x) by ) x – r ) then f) r ) = R Proof:

f)x) = p)x) )x – r)+ R Substitute x = r in both sides f)r) = p)r) )r – r)+ R f) r ) = R

Corollary : Factorization If r is a root of the equation f)x) = 0 then R = 0, f) r) = 0

f) x ) = )x – r) p) x )

The Fundamental Theorem of Algebra The algebraic equation of order n has n roots )complex or real) some of them may be repeated )multiple roots)

If f ) x ) = 0 is an algebraic equation of order n with real coefficients then its complex roots occur as complex conjugate pairs a+ib

a-ib

Horner’s Division Dividing a polynomial of order n by the factor ) x – r ) n

anx + an− 1x

n− 1

+ an− 2 x

= (x − r)(bn− 1x

n− 1

n− 2

+ ⋅ ⋅ ⋅ + a0

+ bn − 2 x

n− 2

+ bn − 3 x

+ ⋅ ⋅ ⋅ + b1x + b0 ) + R

n− 3

n

anx + an− 1x

n− 1

+ an− 2 x

= (x − r)(bn− 1x

n− 1

n− 2

+ ⋅ ⋅ ⋅ + a0

+ bn − 2 x

n− 2

+ bn − 3 x

n− 3

+ ⋅ ⋅ ⋅ + b1x + b0 ) + R bn − 1 x

R

n− 1

+ bn − 2 x

n− 2

+ bn − 3 x

n− 3

Quotient Remainder

+ ⋅ ⋅ ⋅ + b1x + b0

an

= bn-1

→

bn-1 = an

an-1 = bn-2 – r bn-1 →bn-2 = an-1 + r bn-1 an-2 = bn-3 – r bn-2 →bn-3 = an-2 + r bn-2  a 1 = bo – r b1

→

bo = a1 + r b1

a o = R – r bo

→

R = ao + r b o

r an

a n− 1

a n− 2 ...... a 1

r b n− 1 r b n− 2 b n− 1 b n− 2

ao

r b1 r b o

b n− 3 ...... b o

R

Example

Find the quotient and the remainder when dividing f(x) = x3 – 10 x2 + 27 x – 18

by ) x – 2 ) Also find f) 2 )

2 1 − 10 27 − 18 2 1

2 1 − 10 27 − 18 2 − 16 22 1

−8

11

4

2 1 − 10 27 − 18 2 1

−8

Quotient p)x) = x2 – 8 x + 11

Remainder R = 4 = f(2)

Division by the factor (ax − b) f(x) = (ax − b)p(x) + R 1 b R f(x) = (x − )p(x) + a a a f(x) = (x − b a)(ap(x)) + R

Example

Find the quotient and the remainder when dividing

f(x) = 4x by

3

+ x +1

(2 x − 1) a= 2

b=1

f(x) x 1 3 =2 x + + 2 2 2 f(x) x 1 3 2 =2 x +0 x + + 2 2 2 1 2

2

0 1

2

1

1 2 1 2

1

1 2 1 2

1

1 2

2

1 2 1 2

0 1

2

1

1

p(x) 2

p(x) = 2x + x + 1

1 2 1 2

1 R a

R =2

f(x) = 4 x 1 2

4 4

3

+ 0x

2

+ x +1

0

1

1

2

1

1

2

2

2

ap(x)

R

Example

Find the quotient and the remainder when dividing f(x) = x

by

x

2

4

+ 2x

3

− 3x + 5

− 5x + 6

(x − 2)(x − 3)

2 1 2 0 −3

5

2 8 16 26 1 4 8 13 31 3

2

f(x) = (x − 2)(x + 4x + 8x + 13) + 31

3 1 4 8 13 3 21 87 1 7 29 100

f(x) = (x − 2)[(x − 3)

]

(x2 + 7x + 29) + 100 + 31

p(x) = x

2

+ 7x + 29

R(x) = (x − 2)(100) + 31 = 100x - 169