# Polynomial Worksheet

### Solutions

1 State whether the following algebraic expressions are polynomials or not. In the affirmative case, indicate what its degree and independent term are.

1x4 − 3x5 + 2x2 + 5

2 + 7X2 + 2

31 − x4

4

5x3 + x5 + x2

6x − 2x−3 + 8

7

2Write:

1 An ordered polynomial without a constant term.

2 A complete polynomial but not orderly.

3A complete polynomial without an independent term.

4 A polynomial of degree 4, complete with odd coefficients.

3 Given the polynomials:

P(x) = 4x2 − 1

Q(x) = x3 − 3x2 + 6x − 2

R(x) = 6x2 + x + 1

S(x) = 1/2x2 + 4

T(x) = 3/2x2 + 5

U(x) = x2 + 2

Calculate:

1P(x) + Q (x) =

2P(x) − U (x) =

3P(x) + R (x) =

42P(x) − R (x) =

5S(x) + T(x) + U(x) =

6S(x) − T(x) + U(x) =

4 Given the polynomials:

P(x) = x4 − 2x2 − 6x − 1

Q(x) = x3 − 6x2 + 4

R(x) = 2x4 − 2x − 2

Calculate:

P(x) + Q(x) − R(x) =

P(x) + 2 Q(x) − R(x) =

Q(x) + R(x) − P(x)=

5Multiply:

1(x4 − 2x2 + 2) · (x2 − 2x + 3) =

2 (3x2 − 5x) · (2x3 + 4x2 − x + 2) =

3 (2x2 − 5x + 6) · (3x4 − 5x3 − 6x2 + 4x − 3) =

6Divide:

1(x4 − 2x3 − 11x2 + 30x − 20) : (x2 + 3x − 2)

2(x 6 + 5x4 + 3x2 − 2x) : (x2 − x + 3)

3 P(x) = x5 + 2x3 − x − 8         Q(x) = x2 − 2x + 1

7Divide using Ruffini's rule:

1 (x3 + 2x + 70) : (x + 4)

2(x5 − 32) : (x − 2)

3 (x4 − 3x2 + 2 ) : (x −3)

8 Find the remainder of the following divisions:

1(x5 − 2x2 − 3) : (x −1)

2(2x4 − 2x3 + 3x2 + 5x + 10) : (x + 2)

3 ( x4 − 3x2 + 2) :  (x − 3)

9 Indicate which of these divisions are exact:

1(x3 − 5x −1) : (x − 3)

2(x6 − 1) : (x + 1)

3(x4 − 2x3 + x2 + x − 1) : (x − 1)

4(x10 − 1024) : (x + 2)

10Verify the following statements:

1(x3 − 5x −1) has of factor (x − 3)

2(x6 − 1) has of factor (x + 1)

3(x4 − 2x3 + x2 + x − 1) has of factor (x − 1 )

4(x10 − 1024) has of factor (x + 2)

## 1

1State whether the following algebraic expressions are polynomials or not. In the affirmative case, indicate what its degree and independent term are.

1x4 − 3x5 + 2x2 + 5

Degree: 5, independent term: 5.

2 + 7X2 + 2

It is not a polynomial because the literal part of the first monomial is within a root.

31 − x4

Degree: 4, independent term: 1.

4

It is not a polynomial because the exponent of the first monomial is not a natural number.

5x3 + x5 + x2

Degree: 5, independent term: 0.

6x − 2 x−3 + 8

It is not a polynomial because the exponent of the 2nd monomial is not a natural number.

7

Degree: 5, independent term: −7/2.

## 2

Write:

1 An ordered polynomial without a independent term.

3x4 − 2x

2 A complete polynomial but not orderly.

3x − x2 + 5 − 2x3

3A complete polynomial without an independent term.

Impossible

4 A polynomial of degree 4, complete with odd coefficients.

x4 − x3 − x2 + 3x + 5

## 3

Given the polynomials:

P(x) = 4x2 − 1

Q(x) = x3 − 3x2 + 6x − 2

R(x) = 6x2 + x + 1

S(x) = 1/2x2 + 4

T(x) = 3/2x2 + 5

U(x) = x2 + 2

Calculate:

1P(x) + Q (x) =

= (4x2 − 1) + (x3 − 3x2 + 6x − 2) =

= x3 − 3x2 + 4x2 + 6x − 2 − 1 =

= x3 + x2 + 6x − 3

2P(x) − U (x) =

= (4x2 − 1) − (x2 + 2) =

= 4x2 − 1 − x2 − 2 =

= 3x2 − 3

3P(x) + R (x) =

= (4x2 − 1) + (6x2 + x + 1) =

= 4x2 + 6x2 + x − 1 + 1 =

= 10x2 + x

42P(x) − R (x) =

= 2 · (4x2 − 1) − (6x2 + x + 1) =

= 8x2 − 2 − 6x2 − x − 1 =

= 2x2 − x − 3

5S(x) + T(x) + U(x) =

= (1/2 x2 + 4 ) + (3/2 x2 + 5 ) + (x2 + 2) =

= 1/2 x2 + 3/2 x2 + x2 + 4 + 5 + 2 =

= 3x2 + 11

6S(x) − T(x) + U(x) =

= (1/2 x2 + 4) − (3/2 x2 + 5) + (x2 + 2) =

= 1/2 x2 + 4 − 3/2 x2 − 5 + x2 + 2 =

= 1

## 4

Given the polynomials:

P(x) = x4 − 2x2 − 6x − 1

Q(x) = x3 − 6x2 + 4

R(x) = 2x4 − 2 x − 2

Calculate:

P(x) + Q(x) − R(x) =

= (x4 − 2x2 − 6x − 1) + (x3 − 6x2 + 4) − ( 2x4 − 2 x − 2) =

= x4 − 2x2 − 6x − 1 + x3 − 6x2 + 4 − 2x4 + 2x + 2 =

= x4 − 2x4 + x3 − 2x2 − 6x2 − 6x + 2x − 1 + 4 + 2 =

= −x4 + x3 − 8x2 − 4x + 5

P(x) + 2 Q(x) − R(x) =

= (x4 − 2x2 − 6x − 1) + 2 · (x3 − 6x2 + 4) − (2x4 − 2x − 2) =

= x4 − 2x2 − 6x − 1 + 2x3 − 12x2 + 8 − 2x4 + 2x + 2 =

= x4 − 2x4 + 2x3 − 2x2 − 12x2 − 6x + 2x − 1 + 8 + 2 =

= −x4 + 2x3− 14x2 − 4x + 9

Q(x) + R(x) − P(x)=

= (x3 − 6x2 + 4) + (2x4 − 2x − 2) − (x4 − 2x2 − 6x − 1) =

= x3 − 6x2 + 4 + 2x4 −2x − 2 − x4 + 2x2 + 6x + 1=

= 2x4 − x4 + x3 − 6x2 + 2x2 −2x + 6x + 4 − 2 + 1=

= x4 + x3 − 4x2 + 4x + 3

## 5

Multiply:

1(x4 − 2x2 + 2) · (x2 − 2x + 3) =

= x 6 − 2x5 + 3x4 − 2x4 + 4x3 − 6x2 + 2x2 − 4x + 6=

= x 6 − 2x5 − 2x4 + 3x4 + 4x3 + 2x2 − 6x2 − 4x + 6 =

= x 6 −2x5 + x4 + 4x3 − 4x2 − 4x + 6

2 (3x2 − 5x) · (2x3 + 4x2 − x + 2) =

= 6x5 + 12x4 − 3x3 + 6x2 − 10x4 − 20x3 + 5x2 − 10x =

= 6x5 + 12x4 − 10x4 − 3x3 − 20x3 + 6x2 + 5x2 − 10x =

= 6x5 + 2x4 − 23x3 + 11x2 − 10x

3 (2x2 − 5x + 6) · (3x4 − 5 x3 − 6 x2 + 4x − 3) =

= 6x6 − 10x5 − 12x4 + 8x3 − 6x2

− 15x5 + 25x4 + 30x3 − 20x2 + 15x +

+18x4 − 30x3 − 36x2 + 24x − 18 =

= 6x6 − 10x5 − 15x5 − 12x4 + 25x4 + 18x4 +

+8x3 − 30x3 + 30x3 − 6x2− 20x2 − 36x2 + 15x + 24x − 18 =

= 6x6 − 25x5 + 31x4 + 8x3 − 62x2 + 39x − 18

## 6

Divide:

1(x4 − 2x3 − 11x2 + 30x − 20) : (x2 + 3x − 2)

2(x 6+ 5x4 + 3x2 − 2x) : (x2 − x + 3)

3 P(x) = x5 + 2x3 − x − 8         Q(x) = x2 − 2x + 1

## 7

Divide using Ruffini's rule:

1 (x3 + 2x +70) : (x + 4)

2(x5 − 32) : (x − 2)

C(x) = x4 + 2x3 + 4x2 + 8x + 16 R = 0

3 (x4 −3x2 +2) : (x −3)

C(x) = x3 + 3x2 + 6x +18 R = 56

## 8

Find the remainder of the following divisions:

1(x5 − 2x2 − 3) : (x −1)

Apply the remainder theorem:

R(1) = 15 − 2 · 12 − 3 = −4

2(2x4 − 2x3 + 3x2 + 5x +10) : (x + 2)

Apply the remainder theorem:

R(−2) = 2 · (−2)4 − 2 · (−2)3 + 3 · (−2)2 + 5 · (−2) +10 =

= 32 + 16 + 12 − 10 + 10 = 60

3 (x4 − 3x2 +2) :  ( x − 3)

Apply the remainder theorem:

P(3) = 34 − 3 · 32 + 2 = 81 − 27 + 2 = 56

## 9

Indicate which of these divisions are exact:

1(x3 − 5x −1) : (x − 3)

Apply the remainder theorem:

P(3) = 33 − 5 · 3 − 1 = 27 − 15 − 1 ≠ 0

It is not exact.

2(x6 − 1) : (x + 1)

Apply the remainder theorem:

P(−1)= (−1)6 − 1 = 0

Exact.

3(x4 − 2x3 + x2 + x − 1) : (x − 1)

Apply the remainder theorem:

P(1) = 14 − 2 · 13 + 1 2 + 1 − 1 = 1 − 2 +1 +1 − 1 = 0

Exact.

4(x10 − 1024) : (x + 2)

Apply the remainder theorem:

P(−2) = (−2)10 − 1024 = 1024 − 1024 = 0

Exact.

## 10

Verify the following statements:

1(x3 − 5x −1) has of factor (x − 3)

Apply the factor theorem:

(x3 − 5x −1) is divisible by (x − 3) if and only if P(x = 3) = 0.

P(3) = 33 − 5 · 3 − 1 = 27 − 15 − 1 ≠ 0

(x − 3) is not a factor.

2(x6 − 1) has of factor (x + 1)

Apply the factor theorem:

(x6 − 1) is divisible by (x + 1) if and only if P(x = − 1) = 0.

P(−1) = (−1)6 − 1 = 0

(x + 1) is a factor.

3(x4 − 2x3 + x2 + x − 1) has of factor (x − 1)

Apply the factor theorem:

(x4 − 2x3 + x2 + x − 1) is divisible by (x − 1) if and only if P(x = 1) = 0.

P(1) = 14 − 2 · 13 + 1 2 + 1 − 1 = 1 − 2 +1 +1 − 1 = 0

(x − 1) is a factor.

4(x10 − 1024) has of factor (x + 2)

Apply the factor theorem:

(x10 − 1024) is divisible by (x + 2) if and only if P(x = −2) = 0.

P(−2) = (−2)10 − 1024 = 1024 − 1024 = 0

(x + 2) is a factor.