Exercise 1

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

x4 − 3x5 + 2x² + 5

2  \sqrt {x} + 7x ^ 2 + 2

1 − x4

4  \frac {2}{x ^4} - x - 7

x³ + x5 + x²

x − 2x−3 + 8

7  x ^ 3 - x - \frac {7}{2}

Exercise 2

Write:

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.

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) =

2P(x) − U (x) =

3P(x) + R (x) =

42P(x) − R (x) =

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

6S(x) − T(x) + U(x) =

Exercise 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)=

Exercise 5

Multiply:

1(x4 − 2x² + 2) · (x² − 2x + 3) =

2 (3x² − 5x) · (2x³ + 4x² − x + 2) =

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

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

Exercise 7

Divide using Ruffini's rule:

1 (x³ + 2x + 70) : (x + 4)

2(x5 − 32) : (x − 2)

3 (x4 − 3x² + 2 ) : (x −3)

Exercise 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)

Exercise 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)

Exercise 10

Verify 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)

 

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:

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

 

\begin{array}{ccccccccccc} & & &  x^2 & -5 x & +6 &&\\ \cline{3-10} \multicolumn{2}{r}{x^2 +3x - 2 \surd} & x^4 &-2x^3 & -11x^2&+30x &-20 && \\ & &x^4&+3x^3&-2x^2 & & & \\ \cline{3-7} & & &-5x^3&-9x^2 &+30x &-20 & & \\ & & &-5x^3&-15x^2&+10x& & & \\ \cline{4-8} & & & & 6x^2&+20x &-20&& & \\ & & & &6x^2& +18x& -12& & \\ \cline{5-9} & & & & &2x &-8& \\ \end{array}

2(x 6+ 5x4 + 3x² − 2x) : (x² − x + 3)

\begin{array}{ccccccccccc} & & &  x^4 & + x ^3 & +3x ^2 & -6&\\ \cline{3-10} \multicolumn{2}{r}{x^2 -x + 3 \surd} & x^6 &+0x^5 & +5x^4&+0x ^3 &+3x ^2& -2x &+0 && \\ & &x^6&-x^5&+3x^4 & & & \\ \cline{3-7} & & &x^5&+2x^4 &0x ^3 &+3x ^2& -2x &+0 & & \\ & & &x^5&-x^4&+3x ^ 3& & & \\ \cline{4-8} & & & & 3x^4&-3x ^3 &+3x ^2&-2x &+0&& & \\ & & & &3x^4&-3x ^3 &+9x ^2 & & \\ \cline{5-9} & & & &&& -6x^2&-2x &+0&&& & \\ & & & &&&-6x ^2&+6x&-18 & & \\ \cline{5-9} & & & & &&&-8x &+18& \\ \end{array}

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

 

\begin{array}{ccccccccccc} & & &  x^3 & + 2x ^2 & +5x  &+8&\\ \cline{3-10} \multicolumn{2}{r}{x^2 -2x + 1 \surd} & x^5 &+0x^4 & +2x^3&-x &-8&  && \\ & &x^5&-2x^4&+x^3 & & & \\ \cline{3-7} & & &2x^4&+x^3 &0x ^2 &-x & -8 & & \\ & & &2x^4&-4x^3&+2x^2& & & \\ \cline{4-8} & & & & 5x^3&-2x ^2&-x &-8 && & \\ & & & &5x^3&-10x ^2&+5x  & & \\ \cline{5-9} & & & && 8x^2&-6x &-8&&& & \\ & & & &&8x ^2&-16x&+8& & \\ \cline{5-9} & & & & &&10x &-16& \\ \end{array}

 

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.

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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