Polynomial Worksheet

Solution of exercise 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.

Solution of exercise 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

Solution of exercise 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

Solution of exercise 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

Solution of exercise 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

Solution of exercise 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

Solution of exercise 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

Solution of exercise 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

Solution of exercise 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.

Solution of exercise 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.