Polynomials

POLYNOMIALS A. Look at the algebra form below : 1. 2. 3.

3X 2X 6X

+5 X 3 −2 −7 X +4 5 −X 4 + 3 X 3 − 6 X 4

3

2

+10 X −1

Algebra form above are called POLYNOMIALS Degree of polynomials is the index of the highest power. Thus in the expression ( 1 ) above the polynomial is of degree 4 while in ( 2 ) the polynomial is of degree 3 and in ( 3 ) the polynomial is of degree 5. In a polynomial the power of variable is always a positive integer. an X n +a( n −1) X n −1 +a( n −2 ) X n −2 + . . . + a2 X 2 + a1 X +ao , an ≠ 0

is called the

polynomials in X of degree n Exercise 1 Find the degree , variable , coefficient of polynomials below : 1. 6 X 4 + X 3 +3 X 2 −6 X −1 2. 7 Y 7 +5Y 6 + 3Y 3 − Y 2 +3 Y −1 3. 3 a 8 −a 4 + 3a 2 − a +1 4. b 6 +2b 4 − 3b 3 − 6 b 2 +13b −1 5. 4 X 5 −1 6. 6 z 15 −z 14 + 3 z 10 − 6 z 4 +10 z −1 7. 6 +b −b 4 + 3b 5 − 11b 7 +10 b 9 8. 2 X 4Y 3 + 5 X 3Y 2 +3 X 2Y −6 X −1 9. 6a 5 b 6 + a 3 b 4 +2 a 2 b −7 a −14 10. 2a 4 b 4 −2 a 3 b 3 +4 ab 2 −5 b +11 B. Value of polynomials If a polynomial is stated by f(x) and X is replaced by a number ( a ) thus f ( a ) is value of f(x) If f ( x ) = 3 X 2 +7 X +1 , so value of f ( 1 ) = 3 (1) 2 +7 (1) +1 = 11

Exercise 1

Find out value of each polynomials below 1. 2. 3. 4. 5. 6. 7. 8.

f ( x ) = 2 X 4 + X 3 +−3 X 2 −6 X −1 , if x = 2 f ( x ) = −3 X 5 + 7 X 3 −4 X 2 + X −5 , if x = - 1 f ( x ) = 5 X 2 −6 X + 2 , if x = 4 f ( X ) = X 6 −2 X 4 +3 X 2 − X +2 , if x = 3 f ( a ) = 5a 4 −2a 3 +4a 2 −6a −5 , if a = -2 f ( b ) = 3b 5 + b 3 −2b 2 −5 b −4 , if b = - 1 f ( x , y ) = 2 X 4Y 3 + 5 X 3Y 2 +3 X 2Y −6 X −1 , if x = -1 , y = 1 f ( a, b ) = 6a 5 b 6 + a 3 b 4 +2 a 2 b −7 a −14 , if a = 2, b = - 1

C. Operation on Polynomials 1. Addition 2. Subtraction 3. Multiplication Addition n n −1 +a ( n −2 ) X n −2 +. . . +a 0 , a n ≠ 0 and If f ( x ) = a n X +a ( n −1) X h ( x ) = bm X

m

+b( m −1) X

f ( x ) + g ( x ) = bm X

m

m −1

+b( m −2 ) X

m −2

+. . . +b0 , bm ≠ 0 and m > n thus + . . . + ( a 2 + b2 ) X 2 + ( a1 +b1 ) X +( a 0 + b0 )

Example Add

( 6X

4

+X

3

+3 X

2

−6 X −1 ) and ( 5 X

4

− 3 X 3 +6 X

2

+ 7 X +14 )

Solution 6 X 4 + X 3 +3 X 2 −6 X −1 5 X 4 −3 X 3 +6 X 2 + 7 X +14 11X 4 − 2 X 3 +9 X 2 + X +13

+

Subtraction n n −1 +a ( n −2 ) X n −2 +. . . +a 0 , a n ≠ 0 and If f ( x ) = a n X +a ( n −1) X h ( x ) = bm X

m

+b( m −1) X

m f ( x ) - g ( x ) = − bm X

m −1

+b( m −2 ) X

m −2

+. . . +b0 , bm ≠ 0 and m > n thus + . . . + ( a 2 − b2 ) X 2 + ( a1 −b1 ) X +( a0 − b0 )

Example If p ( x) = 6 X 4 + X 3 +3 X 2 −6 X −1 ) and q ( x ) = 5 X find p ( x ) −q ( x ) Solution 6X

4

+X

3

+3 X

2

−6 X −1

4

− X 3 +6 X 2 + 7 X +14

5 X 4 − 3 X 3 +6 X 2 + 7 X +14 X 4 + 4 X 3 −3 X 2 −13 X −15

Find

-

q ( x ) −p ( x )

Solution 5 X 4 − 3 X 3 +6 X 2 + 7 X +14 6 X 4 + X 3 +3 X 2 −6 X −1 − X 4 − 4 X 3 +3 X 2 +13 X +15

-

Multiplication To multiply two polynomials, P ( x ) , Q ( x ) , each term of P ( x ) is multiplied by each term of Q(X) Examine followings 1. Given p ( x) = a x n and q ( x) =b x m we have) p ( x ) . q (x) = a x n b x m . = ab x m +n m m −1 +b( m−2 ) X m −2 +. . . +b0 2. Given p ( x) = a x n and q ( x ) = bm X +b( m −1) X n m m −1 +. . . + b0 ) We have p ( x) . q (x) = a x ( bm x + b( m−1) x n n −1 +a( n −2 ) X n −2 +. . . +a0 and 3. given p ( x ) = a n X +a( n−1) X

q ( x ) =bm X m +b( m−1) X m −1 +b( m−2 ) X m −2 +. . . +b0 we have

Example 1.

p ( x ) =2 X 3 , q ( x ) = 4 x 5 , thus p ( x ).q ( x ) =2 x3 . 4 x 5 = 8 x8

2.

3.

p ( x ) =6 X 2 , q ( x ) = 5 X 4 + 7 X + 8 , thus p ( x ) . q ( x ) = 6 X 2 . ( 5 X 4 + 7 X + 8)

= 30 x 6 +42 x 3 +48 x 2 p ( x) = 3 X 2 −6 X −1 , q ( x ) = 7 X +14 , thus p ( x ) . q ( x ) = ( 3 X 2 −6 X −1) . ( 7 X +14 ) = 21 x 3 + 42 x 2 −42 x 2 −84 x −14 = 21 x 3 −84 x −14

Exercise 2 If p ( x ) =1 +2 x −4 x 2 , q ( x ) =3 x −1 , r ( x ) =3 −2 x 2 , perform the followings 1. p ( x ) +q ( x ) −r ( x ) 2. p ( x ) +q ( x ) +r ( x ) 3. p ( x ) −q ( x ) −r ( x ) 4. 3 p ( x ) +q ( x ) 5. p ( x ) . q ( x ) . r ( x )

6. p ( x ). r ( x ) 7. [ p ( x ) ] 2 +q ( x ) 8. p ( x ) .2 q ( x ) 9. 2 p ( x ) +( q ( x ) . r ( x) ) 10. p ( x ) . ( q ( x ) + r ( x ) )

Exercise 3 State the matter below into

an X n +a( n −1) X n −1 +a( n −2 ) X n −2 +. . . +a0

( x −1 ) 2 ( x +4 )

1. 2. 3. 4. 5. 6. 7.

( x −5 ) 2 − ( 3 x 2 + x −1 ) ( x 2 +2 ) 2 + ( x −4 ) 3 ( x 3 +2 ) 2 ( x −4 ) ( x +2 ) ( x 2 +3 x −4 ) ( x +1) 3 − ( x 2 −4) ( 2 x 2 −x −5 ) −( x +2 ) 3

D. Equality of Polynomials f ( x ) = a n X n +a( n −1) X n −1 +a( n −2 ) X n −2 +. . . +a0 g ( x ) = bm X m +b( m−1) X m −1 +b( m −2 ) X m −2 +. . . +b0 f ( x ) ≡g ( x ) if

m = n, an = bm , a( n −1 ) = b ( m −1) , ........., a0 = b 0

Example 1. Find out the value of a if Solution

( x 2 −3 x +14 ) ≡( x −1 )( x −2) +3a

( x 2 −3 x +14 ) ≡( x −1 )( x −2) +3a

( x 2 −3 x +14 ) ≡( x 2 −3 x +2 +3a ) ( x 2 −3 x +14 ) ≡ x 2 −3 x +( 2 +3a )

Look at the formula above Thus 14 = 2 + 3a 12 = 3a 4 =a Thus value of a on ( x 2 −3 x +14 ) ≡( x −1 )( x −2) +3a is a = 4 2. Find out value of m and n 3x + 4 m n ≡ + x +1 x −2 x − x −2 2

Solution

3x + 4 m n ≡ + x +1 x −2 x − x −2 3x + 4 m ( x −2 ) n ( x +1) ⇔ ≡ + 2 ( x −1)( x − 2) x − x − 2 ( x +1)( x − 2) 2

⇔

⇔

⇔

3x + 4 m ( x − 2 ) + n ( x +1) ≡ ( x +1)( x − 2) x −x −2 2

3x + 4 m x − 2m + n x + n ≡ ( x +1)( x − 2) x −x −2 3x + 4 ( m + n ) x + ( − 2m + n ) ≡ ( x +1)( x − 2) x2 − x − 2 2

m + n =3 −2m +n = 4

so

Use elimination or substitute is got m =

−1 1 and n = 3 3 3

Exercise 4 Find out the value of a 1. ( x +1 )( x +3) −2a ≡x 2 +4 x −1 2. ( x 2 +1 ) ( x −5 ) +3a ≡x 3 −5 x 2 +x +16 3. x 4 +2 x 3 + x 2 +4 x −3 ≡ ( x 2 +2)( x 2 +2 x −1) + a Find out the value of m , n 1. 2 − nx −mx 2 −x 3 ≡( 1 −x )( x 2 −3 x +2) 2. 10 −18 x +7 x 2 −x 3 ≡( 3 −x ) 2 (1 − x ) +mx +n Find out the value of a, b or c a

b

4x

1. x −2 + x +2 ≡ x 2 −4 3 x +4

a

b

2. x 2 − x −2 ≡ x − 2 + x +1 2 x 2 +x +2 bx +c a ≡ − 2 3 1 −x x +x +1 1 −x

3.

E. Division on Polynomials If f(x) and g (x ) are polynomials and the degree of f(x) > the degree g (x ) > 1. There are polynomial h(x) and s(x) are fulfill f (x ) = g(x) . h ( x ) + s(x) f (x ) = g(x) . h ( x ) + s(x) f(x) , g( X), h(x),s(x) are called dividend , divisor, quotient, remainder Example E . 1 Polynomials by ( x – a ) 1. Divide f ( x ) =x 2 +4 x −8 by h( x ) =x −1 Solution x x

− 1 x

2

+ 5

+ 4x − 8

x2 − x 5x − 8 5x − 5 − 3

Hence x 2 +4 x −8 =( x −1)( x +5) −3 x 2 +4 x −8 is called dividend x −1

is called divisor

x +5

is called quotient

-3

is remainder

2. Divide f ( x ) = x 3 +2 x 2 + 3 x −5 by h( x ) =x −2 Solution x

− 2 x

3

x 2 + 4x + 11 + 2x 2 + 3x − 5

x3 − 2x 2 4x2 + 3x 4x2 − 8x 1 1x − 5

− −

11 x − 22 − 17

Hence x 3 + 2 x 2 + 3 x −5 ≡ ( x − 2)( x 2 +4 x +11) +17

x

3

+2 x 2 +3 x −5

x −2

is called dividend is called divisor

x 2 +4 x +11

is called quotient

17

is remainder

Exercise 5 Find out the quotient and remainder at division of polynomials below : 1. 3 x 2 −2 x +1 : x + 3 2. 2 x 3 −5 x 2 +4 x +3 : x +1 3. x 4 +3 x 3 +4 x 2 −x +1 : x −1 4. 4 x 4 −1 : 2 x −1 5. 3 x 3 − 2 x 2 + 6 : x − 2 6. 5 x 5 − 2 x 3 + 4 x −1 : x 3 + 2 x −1 7. x 5 −3 x 3 +4 x 2 −1 : x 2 − x +3 8. − 2 x 5 + 6 x 4 − 3 x 3 + 6 : x 2 − 2 x + 3 9. 4 x 3 −3 x 2 −6 x −5 : x 2 +x Find p (x ) in each equation below : 1. 3 x 3 −x 2 +6 ≡ ( x −2 ) . p ( x ) +26

2. 2 x 5 − x 3 +3 x +4 ≡(16 x +64) +( x 2 − x +4). p ( x ) 3. ( x −1 ). p ( x ) +7 ≡ x 3 +4 x 2 − 2 x +4 4. x 4 −2 x 3 +x 2 −5 x +10 ≡ ( x −1) . p ( x ) +5 F. Horner’s Method F . 1 Polynomial is divided by ( x – a ) 1. a. f ( x) = ( x 3 +6 x 2 +3 x −15 ) divided by ( x +3 ) Solution Method 1 x 2 + 3x − 6 x + 3 x3 + 6 x 2 + 3 x −15 x3 + 3 x2 3 x2 + 3 x 3x 2 + 9 x − 6 x − 15

− −

− 6 x −18 − 3

x

3

is called dividend

+6 x 2 +3 x −15

x −3

is called divisor is called quotient

x 2 +3 x −6

3

is remainder

Method 2 Horner method b. f ( x) = ( x 3 +6 x 2 +3 x −15 ) divided by ( x – 3 ) 3

1

1

6

3

-15

-3

-9

18

3

-6

3 = remainder

is called dividend

+6 x 2 +3 x −15

3

x

x −3

is called divisor is called quotient is remainder

x 2 +3 x −6

3

2. f ( x) = (2 x 4 −5 x 2 +2 x −2) divided by ( x −1 ) Method 1 2x 3 + 2 x 2 − 3 x −1 x −1 2 x 4 − 5 x 2 +2 x −2 2 x 4 −2 x 3 2 x 3 −5 x 2

−

2 x3 − 2 x2 − 3x 2 + 2x

−

− 3x 2 +3 x − − x −2 −x +1 − −3

2x

4

is called dividend

−5 x 2 +2 x − 2

x −1

is called divisor

2 x 3 +2 x 2 −3 x −1

is called quotient

-3

is remainder

Method 2 1

2

2

2x

4

0

-5

2

-2

2

2

-3

-1

2

-3

-1

-3

−5 x 2 +2 x − 2

x −1

is called dividend is called divisor

is called quotient

2 x 3 +2 x 2 −3 x −1

-3

is remainder

F . 2 Polynomial is divided by ( ax – b) Polynomials f(x) divided by ( ax – b ) b a

Thus f ( x ) = h( x )( x − ) + s ( x ) h ( x) = quotient s ( x) =remainders

: divide 2 x 3 + 9 x 2 −6 x +4 by 2 x +1

Example x 2 + 4x − 5 2x +1 2x 3 + 9 x 2 − 6x + 4 2x 3 + x 2 8x 2 − 6x

−

8x 2 + 4x − 10 x + 4

−

− 10 x 2 − 5 − 9

Horner’s Method -½

2

9

-6

4

-1

-4

5

8

- 10

9

2 2x

3

is called dividend

+9 x 2 −6 x +2 2 x +1

is called divisor

2 x 2 +8 x 2 −10 = x 2 +4 x 2 −5 2

9

is called quotient is remainder

F . 3 Polynomial is divided by ( x – a)( x – b ) Example Find the quotient and remainders when 2 x 4 + 6 x 3 − 5 x 2 + 3 x − 7 is divided by ( x – 1 )( x + 3)

Solution 1 2 x2 + 2 x − 3 x 2 + 2 x − 3 2x 4 + 6 x 3 − 5x 2 + 3x − 7 2x 4 + 4 x 3 − 6x 2 − 2 x3 + x2 + 3 x 2 x3 + 4 x 2 − 6 x − − 3x 2 + 9 x − 7 − 3x 2 − 6x + 9 15 x −16

Solution 2 Horner ‘s Method 1

2

2 -3 2

6

-5

3

-7

2

8

3

6

8

3

6

-1 = S1

-6

-6

9

2

-3

15 = s 2

S(x) = ( x – 1 ) s 2 + S1 = ( x – 1 ) 15 + ( - 1 ) = (15 x – 15 – 1 ) =15 x – 16 Or -3

2

2 1 2

6

-5

3

-7

-6

0

15

-54

0

-5

18

- 61 = S1

2

2

-3

2

-3

15 = S 2

S ( x ) = ( x + 3 ) S 2 + S1 = ( x +3 ) .15 – 61 = 15x + 45 – 61 = 15 x – 16 Exercise 6 Find the remainders and quotient P (x) divided by q ( x ) using Horner ‘s Method 1. x 4 +3 x 3 +4 x 2 −x +1 : x −1 2. 2 x 4 +6 x 3 4 x 2 −6 x +1 : x −3

3. x 4 −2 x 3 +5 x 2 + 4 x +3 : x + 2 4. 4 x 4 + x 2 −3 x − 7 : x +1 5. x 4 +4 x 3 +8 x 2 −4 x +1 : 2 x −1 6. 3 x 5 −4 x 4 +4 x 3 −7 x 2 +1 : 3 x −1 7. 2 x 4 +4 x 3 +8 x 2 −4 x +1 : x 2 −x −2 8. 2 x 4 − 3 x 3 − 5 x 2 + 6 x + 4 : ( x 2 + 3 x + 2) 9. x 5 +3x 3 −6 x 2 + x −3 : ( x 2 −2 x −8) 10. x 6 −4 x 2 −x +3 : ( x 2 +3 x −4 )

If f(x) and g(x) are polynomials and degree of f( x ) > g(x) than : 1. Degree of g(x) is 1 ,the remainders of f(x) divided by g(x) is s(x), and degree of s(x) is zero 2. Degree of g(x) is 2 ,the remainders of f(x) divided by g(x) is s(x), and degree of s(x) is one 3. Degree of g(x) is 3 ,the remainders of f(x) divided by g(x) is s(x), and degree of s(x) is two 4.

G. The Remainder Theorem Theorem : If polynomial p(x) is divided by ( x – a ) the remainder is S = P ( a ) p ( x ) =( x −a ) . h ( x ) + s ( x −a ) =0 ⇒x =a Substitute a for p ( x ) =( x −a ) . h ( x ) + s p ( a ) =( a −a ) . h ( a ) + s ⇒ p(a)= s

Corollaries 1. If p(x) is divisible by x – a , then the remainders is 0, that is P ( a ) = 0 2. If p (x) is divisible by ( x – a ), ( x – b ), ( x – c ) than is divisible by ( x – a ), ( x – b ), ( x – c ) separately 3. If p ( x ) is divisible (x – a ) then ( x – a ) is factor of p ( x ) and x = a is root of p ( x ) Example 1. Find the remainders if p ( x ) = 2 x 4 − 2 x 3 +5 x 2 −4 x +5 divided by ( x – 1 ) Solution

1

2

-2

5

-4

5

2

2

0

5

1

0

5

1

6

=S=p(1)

2. Find a if p ( x ) = x 4 −x 2 + ax +2 divisible by ( x + 2 ) Solution -2

1

1

So

0

-1

a

2

-2

4

-6

12 – 2a

-2

3

- 6 +a

14 – 2a = S = 0

12 – 2a = 0 a =6

3. If the remainders is 3x – 5, when p ( x ) = x 3 −mx 2 +2 x +4 is divided by x 2 −1 . Find of m. Solution If p ( x ) = x 3 −mx 2 +2 x +4 is divided by x 2 −1 . so 1. x 3 −mx 2 +2 x +4 = ( x −1)( x +1) h( x) +3x −5

13 −m(1) 2 +2.1 +4 =(1 −1 )(1 +1) h(1) + 3.1 − 5 7 −m = −2 m =9

Or 2.

x 3 −mx 2 +2 x +4 = ( x −1)( x +1) h( x ) +3 x −5

( −1) 3 −m( −1) 2 +2.( −1) +4 =( −1 −1 )( −1 +1) h( −1) + 3.( −1 ) − 5 1 −m = −8 m =9

So value of m is 9 4. f(x) is polynomials, 5 is remainder if f(x) divided by ( x – 2).and – 10 is remainders if f(x) divided by ( x + 3). Find remainder if f(x) divided by ( x 2 +x −6 ) Solution f(x) = ( x 2 +x −6 ) h(x) + s(x) f(x) = ( x 2 +x −6 ) h(x) + ( ax + b )

f(x) f(2) 5 5

= ( x +3)( x −2 ) h(x) + (ax + b ) = ( 2 +3)(2 −2 ) h(2) + (2a + b ) = 0 + (2a + b ) = 2a + b ……………… ⇒ ( i )

f(-3) = ( −3 +3)(−3 −2 ) h(-3) + (-3a + b ) f(-3) = 0 + (-3a + b ) - 10 = - 3a + b …………………. ⇒

( ii )

( i ) and ( ii ) 5 = 2a + b - 10 = - 3a + b

thus a = 3 , b = - 1

2 Thus the remainders of f(x) when divided by ( x +x −6 ) is 3 x - 1

Exercise 6 Solve the following problems below 1. When p ( x) =x 3 +3 x 2 − mx +n is divided by ( x – 1 ) and ( x – 2 ) the remainders are 3 and 9 respectively. Find value of m and n 2. When p ( x ) =3 x 3 +2 x 2 − 4 x +n is divided by ( x + 1 ) the remainder is 7. find the value of n 3. x 3 − 4 x 2 + m x + n ≡ ( x 2 − 3x + 2)h( x) + ( 6 − 2 x) . Find the value of m and n 4. If the remainders are 3 and 21 when p(x) is divided by x – 1 and x + 2 respectively, Find the remainders when p(x) is divided by ( x 2 +x −2 ) 5. f(x) is polynomials. When f(x) divided by ( x 2 − x ) has the remainders 4 – 3x and when f(x) divided by ( x 2 + x ) has the remainders (4 – x ). Find the remainders if f(x) divided by ( x 2 −1 ) 6. If the remainders are – 1, - 12 , 31 when p(x) divided by ( x – 1 ), (x – 2 ), ( x – 3 ) respectively, Find the remainders when p(x) id divided by ( x – 1 ). (x – 2 ). ( x – 3 ). 7. When p ( x ) =ax 5 −2 x 4 + 3 x 3 +4 x 2 +bx −6 divided by ( x – 2 ) , ( x- 3 ) has remainders 12 and 159. Find the remainders when p(x) divided by ( x – 1 )(x – 2 )(x – 3) 8. If f ( x) =19( x 21 −x 8 +2) −15( x17 −4 x 3 + 3 ) find remainder if f(x) divided by a. ( x – 1 ) b. ( x + 1) c. d. 9. fff 10. fff 11. fff 12.

H. kkk I. kkk J. kkk K. kkk L. kkk M. kkk N. kkk O. ;;; P. ;;; Q. ;;; R. ;;; S. ;;; T. ;;; U. ;;; V. ;;; W. Value of polynomial