NOTA POLYNOMIALS

3: POLYNOMIALS

Polynomials are expression such as 2x + 1, x2 – 3x + 4, x3 + 2x – 1.

- General polynomial form

an x n + a x n−1 + a x n−2 + a x n−3 + ......a2 x 2 + a1 x + a0
n−1 n−2 n−3

an, an-1, an-2, an-3, …..a2, a, a0 are coefficients with real numbers, an  0
n is a positive integer

- The degree of the polynomial is the highest power of x in the expression.

Example :

P(x) = 2x7 + 3x5 – x4 + 5x + 1 a polynomial of degree 7

P(x) =x3 – 4x2 + 5x – 7 a polynomial of degree 3

Polynomial of degree 1 – linear
Polynomial of degree 2 – quadratic
Polynomial of degree 3 – cubic
Polynomial of degree 4 – quartic

Operation on polynomials

a) Addition and Subtraction

- collect together terms of the same degree

Example 1:

Given p(x) = 3x4 – x3 + 2x2 + 4x + 1 and q(x) = x4 + 2x3 – x2 + 4, find p(x) + q(x).

Solution :

p(x) + q(x) = (3x4 – x3 + 2x2 + 4x + 1) + (x4 + 2x3 – x2 + 4)
= 3x4+ x4 – x3 + 2x3 + 2x2 – x2 + 4x + 1 + 4
= 3x4+ x4 – x3 + 2x3 + 2x2 – x2 + 4x + 1 + 4
= 4x4 – x3 + x2 + 4x + 5

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Example 2:

Given f(x) = 2x4 + 3x - 1 and g(x) = x4 – 2x3 + 4, find in simplest form each of polynomials
2f(x) - g(x).

Solution :

2f(x) - g(x) = 2(2x4 + 3x - 1) - (x4 – 2x3 + 4)
= 4x4 + 6x - 2 - x4 + 2x3 – 4
= 4x4 - x4 + 2x3 + 6x – 2 – 4
= 3x4 + 2x3 + 6x – 6

b) Multiplication

- Each term of one polynomial is multiplied by each term of the other polynomial.

Example 1:

Given two polynomials p(x) = 2x3 - 3x + 4x and q(x) = x2 + 2x – 5, find p(x)q(x).

Solution :

p(x)q(x) = (2x3 - 3x + 4x ) (x2 + 2x – 5)
= 2x3(x2 + 2x – 5) - 3x(x2 + 2x – 5) + 4(x2 + 2x – 5)
= 2x5 + 4x4 – 10x3 - 3x3 - 6x2 + 15x + 4x2 + 8x – 20
= 2x5 + 4x4 – 13x3 - 2x2 - 23x – 20

Example 2:

Find the coefficient of x3 in the expansion of (2x3 - 3x2 - x + 4) (x – 1).

Solution :

2x3( -1)

(2x3 - 3x2 - x + 4) (x – 1)

-3x2(x)

The term in x3 is -2x3 – 3x3 = -5x3
Therefore, the coefficient of x3 is -5

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Exercises:

1. Simplify (4x3 – 3x2 + x – 6) – (x3 + 4x2 + x – 2)
2. Find the product of x3 – 2x2 + x – 1 and x + 1
3. Find the coefficient of the terms in x2 in the expansion of (x3 + 3x2 + 5x – 1) (x – 2)
4. Find the coefficient of the terms in x2 in the expansion of (x2 + 3x + 5)2

b) Division

- A polynomial can be divided by another by using

i) long division
ii) synthetic division

- Polynomials can be written in the form of :

Polynomials = Divisor x Quotient + Remainder
P(x) = D(x) Q(x) + R

1
3 5 → 5=3x1 +2

3
2

Example :

Find the quotient and remainder when x3 – 5x2 + 2x + 10 is divided by (x – 2)

Method 1

Long division

Divisor x2 − 3x − 4 Quotient
x −2 x3 − 5x2 + 2x +10 Remainder

x3 – 2x2
-3x2 + 2x
-3x2 +6x
-4x + 10
-4x + 8
2

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Therefore the quotient is x2 – 3x – 4 and the remainder is 2.
In the form of P(x) = D(x) × Q(x) + R,

x3 – 5x2 + 2x + 10 = (x – 2)( x2 – 3x – 4)+ 2

Method 2

Synthetic division

- Can only be used if the divisor is a linear polynomial.

Steps :
- Write down all the coefficient of the polynomials
- take 2 as the divisor ( if the divisor is x + 2, take -2)

For x3 – 5x2 + 2x + 10,

1 -5 2 10 add

2 -6 -8 Remainder
2 1 -3 -4 2

multiply

Therefore the quotient is x2 – 3x – 4 and the remainder is 2.

Exercises:

1. By using long division or synthetic division, find the quotient and remainder when
a) x2 + 5x – 6 is divided by x + 2
b) 6x2 – x + 2 is divided 2x + 1
c) x3 – 5x2 + 6x + 8 is divided by x2 – 3x

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REMAINDER THEOREM

- A method of finding the remainder without going through the process of division.

Let us consider finding the remainder when P(x)  3x3 + 2x2 -4x + 9 is divided by x + 2 without
performing long division.

P(x) = D(x) × Q(x) + R
3x3 + 2x2 -4x + 9 = (x + 2) × Q(x) + R

Substitute x = -2

3(-2)3 + 2(-2)2 -4(-2) + 9 = (-2 + 2) × Q(x) + R
3(-8) + 2(4) + 8 + 9 = 0 × Q(x) + R
-24 + 8 + 8 + 9 = R
R=1

When a polynomial P(x) is divided by x – a, the remainder is P(a).

Polynomial is divided by linear function Remark

Example 1: i) If P(x) is divided by x + a = x – (-a), then R = P(-a)
ii) If P(x) is divided by ax – b = a( x − b ), then R = P( b )
Find the remainder when 2x3 – 3x2 – 8x – 3 is divided by
a) x – 3 aa
b) 2x - 1

Solution :
Let p(x) = 2x3 – 3x2 – 8x – 3

a) x = 3
p(3) = 2x3 – 3x2 – 8x – 3
= 2(3)3 – 3(3)2 – 8(3) – 3

= 54 – 27 – 24 – 3

= 0 Therefore, the remainder is 0.

b) x = 1
2

p(1) = 2(1)3 − 3(1)2 − 8(1) − 3
22 2 2

= 1 − 3 −4−3
44

= − 15 Therefore the remainder is − 15
2 2

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Example 2:

When the polynomial x4 + kx2 + 4x – 2 is divided by x + 3, the remainder is 13. Find the value of k.

Solution :

Let P(x) = x4 + kx2 + 4x – 2
P(-3) = 13

(-3)4 + k(-3)2 + 4(-3) – 2 = 13
81 + 9k – 12 – 2 = 14
9k + 67 = 14
9k = 14 – 67
9k = – 54
k = -6

Example 3:

The polynomial 2x3 + ax2 + 3x + b has a remainder -13 when divided by x + 1, and has a remainder
5 when divided by 2x - 1. Find the value of a and b.

Solution :

Let P(x) = 2x3 + ax2 + 3x + b

P(-1) = -13 ---------------------- (1)
2(-1)3 + a(-1)2 + 3(-1) + b = -13

-2 + a – 3 + b = -13
a + b = -8

P(1) = 5 ------------------------(2)
2

2(1)3 + a(1)2 + 3(1) + b = 5
222
1 +1a+3+b=5
44 2
1 + a + 6 + 4b = 20
a + 4b = 13

a + b = -8 -----(1)
a + 4b = 13 -----(2)

-3b = -21
b=7

sub. b = 7 into (1)
a + 7 = -8

a = -15

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Polynomial is divided by quadratic function

Polynomial x3 – 2x2 +5x – 1 is divided by quadratic function x2 – x – 2.

x −1
x2 − x −2 x3 −2x2 + 5x −1

x3 – x2 – 2x
-x2 + 7x – 1
-x2 + x + 2
6x – 3

The remainder is 6x –3

When a polynomial P(x) is divided by quadratic function the remainder is a linear
function in the form Ax + B where A and B are constants.

• Finding the remainder in the form of Ax + B ; P(x) = D(x) × Q(x) + R(x)

Let x3 – 2x2 +5x – 1  (x2 – x – 2) Q(x) + (Ax + B)
x3 – 2x2 +5x – 1  (x + 1)(x – 2) Q(x) + (Ax + B)

For (x + 1)(x – 2) = 0, x = -1 and x = 2.

Subtitute x = -1
(-1)3 – 2(-1)2 +5(-1) – 1 = A(-1) + B

-1 – 2 – 5 – 1 = -A + B
-9 = -A + B --------------------(1)

Subtitute x = 2
(2)3 – 2(2)2 +5(2) – 1 = A(2) + B

8 – 8 + 10 – 1 =2A + B ---------------------(2)

-A + B = -9 ---(1)
2A + B = 9 ---(2)

-3A = -18
A=6

Sub. A = 6 into (1)
-(6) + B = -9
B = -9 + 6
B = -3

Therefore, the remainder is 6x – 3.

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FACTOR THEOREM
- x – a is a factor of P(x) if P(x) is divided by x – a, then the remainder is zero.
- x = a is a root of the equation P(x)=0 and
- a is a zero of P(x)

P(a) = 0 → x – a is a factor of P(x)

Example 1 :
Given that p(x) = x3 – x2 – 3x + 2, show that x – 2 is a factor of p(x).
Solution:
p(2) = (2)3 – (2)2 – 3(2) + 2

= 8–4–6+2
=0
Since p(2) = 0, x – 2 is a factor of p(x).
We can say that 2 is a zero of p(x).

Example 2 :
Prove that 2x – 1 is a factor of P(x) = 2x3 + 3x2 – 8x + 3, and factorise P(x) completely.
Solution:
Substitute x = 1

2
P( 1 ) = 2( 1 )3 + 3( 1 )2 – 8( 1 ) + 3

2 222
= 1+ 3-4+3
44
= 1–1
=0

Since P( 1 ) = 0, 2x – 1 is a factor of P(x).
2

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Use long division to find the other factors.

x2 – 2x - 3
2x − 1 2x3 + 3x2 −8x + 3

2x3 – x2
4x2 – 8x
4x2 – 2x
-6x + 3
-6x + 3

P(x)  (2x – 1)( x2 – 2x – 3 ) ; factorise x2 – 2x – 3
 (2x – 1)( x + 2) ( x – 3 ) ; P(x) is factorised completely

Example 3 :

Polynomial P(x)  2x3 + ax2 – x + b has x + 1 as a factor and leaves a remainder 12 when divided by
x – 3. Find a and b hence find all factors of P(x).

Solution:

P(x)  2x3 + ax2 – x + b

Since x + 1 is a factor
P(-1) = 0

2(-1)3 + a(-1)2 – (-1) + b = 0
-2 + a + 1 + b = 0
-1 + a + b = 0
a + b = 1 ----- (1)

also P(3) = 12 ; from remainder theorem

2(3)3 + a(3)2 – (3) + b = 12

54 + 9a – 3 + b = 12

51 + 9a + b = 12

9a + b = -39 -----(2)

(2) – (1)

9a + b = -39 ---- (2) a +b=1
a + b = 1 ---- (1) -5 + b = 1
8a = -40
a = -5 b=6

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Therefore P(x)  2x3 - 5x2 – x + 6

P(x)  (x + 1) (2x2 – 7x + 6) By using long division
 (x + 1) (x – 2) (2x – 3)

Exercises:

The polynomial 2x3 + ax2 + bx – 2 has a factor x + 2. When the polynomial is divided by 2x – 3, the
remainder is 7. Find the values of a and b. with these values of a and b, find the three linear factors of
the polynomial.

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PARTIAL FRACTION
- Two single algebraic fractions can be combined to become a single compound fraction as
shown:
2 + 3 = 2(4 − x) + 3(3 + x) = x + 17
3 + x 4 − x (3 + x)(4 − x) (3 + x)(4 − x)

- The reverse process of expressing a single compound fraction as a sum of two simpler
fractions is called expressing the algebraic fraction in partial fraction.
x +17 = 2 + 3
(3 + x)(4 − x) 3 + x 4 − x

Partial fraction

Type 1: Denominator with linear factor ax + b

- The partial fraction of the form A
ax + b

Example :

Express x − 11 in partial fractions.
(x + 3)(x − 4)

Solution:  A+B
Let x − 11 x+3 x−4

(x + 3)(x − 4)  A(x − 4) + B(x + 3)
(x + 3)(x − 4)

Therefore, x – 11  A(x – 4) + B(x + 3)

(choose any suitable value of x so that the linear expressions, ax + b = 0)
Substitute x = 4,

-7= B(7)
-7 = 7B
B = -1

Substitute x = -3,
- 14 = A ( -7 )
-14 = - 7A
A =2

11

Hence,
x − 11  2 − 1

(x + 3)(x − 4) x + 3 x − 4

Exercises:

Express 7x − 4 in partial fractions.
x(x − 1)(x + 2)

Type 2: Denominator with a quadratic factor, ax2 + bx + c

- The partial fraction of the form Ax + B
ax 2 + bx + c

Example :

Express 3x + 1 in partial fractions.
(x − 1)(x2 + 1)

Solution:

Let 3x + 1  A 1 + Bx +C
(x − 1)(x2 + 1) x− x2 +1

 A(x2 + 1) + (Bx + C)(x − 1)
(x − 1)(x2 + 1)

Therefore, 3x + 1  A(x2 + 1) + (Bx + C)(x - 1)  Ax2 + Bx2 – Bx + Cx + A - C

Substitute x = 1,
4 = A(2)
4 = 2A
A =2

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Equating coefficient x2,
0 = A x2 + Bx2
0 = A+B
0 = 2+B
B = -2

Equating constant terms,
1 = A- C
C = 2-1
C =1

Hence, 3x + 1  x 2 1 + 1− 2x
(x − 1)(x2 + 1) − x2 +1

Exercises:

Express x−3 + 1) in partial fractions.
(x + 2)(x2

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Type 3: Denominator with a repeated linear factor, (ax + b)2

- The partial fraction of the form A+ B
ax + b (ax + b)2

Example :

Express 2x + 1 in partial fractions.
(x − 1)2 (2x − 5)

Solution:

Let 2x + 1  A + B + C
(x − 1)2 (2x − 5) (x − 1) (x − 1)2 2x −5

 A(x − 1)(2x − 5) + B(2x − 5) + C(x − 1)2
(x − 1)2 (2x − 5)

Therefore 2x + 1 = A(x – 1)(2x – 5) + B(2x – 5) + C(x + 1)2

Substitute x = -1, Substitute x = 5 , Equating coefficient x2,
-1 = - 7B 2 0 = 2A + C
B= 1 6 = 49 C A = - 12
7 4 49
C = 24
49

Hence, 2x + 1  −12 + 1 + 24
(x − 1)2 (2x − 5) 49(x − 1) 7(x − 1)2 49(2x − 5)

Exercises:

Express x−1 in partial fractions.
(x + 1)(x + 2)2

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Type 4 : Degree of numerator is the same or higher than the denominator.

- Long division is used so that the degree of the numerator becomes less than that of the
denominator.

Example :
Express 3x2 − 2x − 7 in partial fractions.

(x − 2)(x + 1)

Solution:  3x2 − 2x − 7
3x2 − 2x − 7 x2 −x−2
(x − 2)(x + 1)

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x2 − x −2 3x2x −2x − 7

3x2 − 3x − 6

x −1

From long division,

3x2 − 2x − 7  3 + x 2 x−1 2
(x − 2)(x + 1) −x−

 3+ x−1
(x − 2)(x + 1)

Let x − 1  A + B
(x − 2)(x + 1) x−2 x+1

x – 1  A(x + 1) + B(x – 2)

Substitute x = -1

-2 = -3B
B= 2

3

Substitute x = 2

1 = 3B

A= 1
3

Hence 3x2 − 2x − 7  3 + 1 + 2
(x − 2)(x + 1) 3(x − 2) 3(x + 1)

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