4 eso academics - unit 03 - polynomials. algebraic fractions

Unit 03 November

1. MONOMIALS AND POLYNOMIALS.

1.1. MONOMIALS.

A Monomial is an Algebraic Expression containing one Term which may be a

number, a Variable or a product of numbers and variables, with no negative or

fractional exponents. (Mono implies one and the ending nomial is Greek for part).

2a3

; 5𝑥𝑥; −2𝑦𝑦; 450𝑥𝑥2𝑧𝑧 are monomials

The number is called Coefficient and the variables are called Literal Part. If the

literal part of a monomial has only one letter, then the Degree is the exponent of the

letter. If the literal part of a monomial has more than one letter, then the degree is the

addition of the exponents of the letters.

The degree of −5𝑥𝑥3 is 3

The degree of 2𝑥𝑥2𝑦𝑦3𝑧𝑧 is 2 + 3 + 1 = 6

MATH VOCABULARY: Monomial, Algebraic Expression, Term, Variable, Coefficient,

Literal Part, Degree, Polynomial.

1.2. ADDITION AND SUBTRACTION OF MONOMIALS.

You can add monomials only if they have the same literal part (they are also

called like terms). In this case, you add the coefficients and leave the same literal part.

Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.1

Unit 03 November

3𝑥𝑥 + 5𝑥𝑥 = 8𝑥𝑥

3𝑥𝑥 − 2𝑥𝑥2 You cannot add the terms because they have different literal part.

1.3. MULTIPLICATION AND DIVISION OF MONOMIALS.

If you want to multiply two or more monomials, you just have to multiply the

coefficients, and add the exponents of the equal letters.

2𝑥𝑥7 ∙ 3𝑥𝑥3 = (2 ∙ 3) ∙ 𝑥𝑥7+3 = 6𝑥𝑥10

(−2𝑥𝑥𝑦𝑦2𝑧𝑧) ∙ (5𝑥𝑥2𝑧𝑧3) = �(−2) ∙ 5� ∙ (𝑥𝑥1+2) ∙ (𝑦𝑦2+0) ∙ (𝑧𝑧1+3) = −10𝑥𝑥3𝑦𝑦3𝑧𝑧4

If you want to divide a monomial by a monomial of the same or lower degree,

you just have to divide the coefficients, and subtract the exponents of the equal

letters.

10𝑥𝑥5 ÷ 2𝑥𝑥2 = (10 ÷ 2) ∙ (𝑥𝑥5−2) = 5𝑥𝑥3

(12𝑎𝑎2𝑏𝑏) ÷ (3𝑎𝑎) = (12 ÷ 3) ∙ (𝑎𝑎2−1) ∙ (𝑏𝑏1−0) = 4𝑎𝑎𝑏𝑏

1.4. POLYNOMIALS.

A Polynomial is the addition or subtraction of two or more monomials (which

are called Terms). If there are two monomials, it is called a Binomial, if there are three

monomials, it is called a Trinomial. The Degree of the polynomial is the highest degree

of the terms that it contains.


Unit 03 November

You usually write polynomials with the terms in “Decreasing” order of exponents. We

say that a polynomial is Complete if it has terms of every exponent from the degree of the

polynomial until you get down to the Constant Term.

Polynomials are also sometimes named for their degree:

MATH VOCABULARY: Binomial, Trinomial, To Decrease, Constant Term, Quadratic,

Cubic, Quartic.


Unit 03 November

1.5. EVALUATING POLYNOMIALS.

“Evaluating” a polynomial 𝑷𝑷(𝒙𝒙) is calculating its numerical value at a given

value of the variable: 𝒙𝒙 = 𝒂𝒂. You must substitute the variable 𝒙𝒙 for the value 𝒂𝒂, and

calculate the value of the polynomial 𝑷𝑷(𝒂𝒂).

Evaluate 𝑃𝑃(𝑥𝑥) = 𝑥𝑥4 − 3𝑥𝑥2 + 𝑥𝑥 + 1 at 𝑥𝑥 = 2

𝑃𝑃(2) = 24 − 3 ∙ 22 + 2 + 1 = 16 − 12 + 2 + 1 = 7

MATH VOCABULARY: Numerical Value.

1.6. ADDING AND SUBTRACTING POLYNOMIALS.

When adding or subtracting polynomials you must add or subtract each like

term of the polynomial, that is, monomials that have the same literal part. (You must

use what you know about the addition of monomials).

If 𝑃𝑃(𝑥𝑥) = 𝑥𝑥2 + 3𝑥𝑥 − 4;𝑄𝑄(𝑥𝑥) = 𝑥𝑥3 + 2𝑥𝑥 + 1 and 𝑅𝑅(𝑥𝑥) = −𝑥𝑥 + 3

Find 𝑃𝑃(𝑥𝑥) + 𝑄𝑄(𝑥𝑥)− 𝑅𝑅(𝑥𝑥) = 𝑆𝑆(𝑥𝑥)

𝑆𝑆(𝑥𝑥) = (𝑥𝑥2 + 3𝑥𝑥 − 4) + (𝑥𝑥3 + 2𝑥𝑥 + 1)− (−𝑥𝑥 + 3) =

= 𝑥𝑥3 + 𝑥𝑥2 + �3𝑥𝑥 + 2𝑥𝑥 − (−𝑥𝑥)� + �(−4) + 1 − 3� = 𝑥𝑥3 + 𝑥𝑥2 + 6𝑥𝑥 − 6

1.7. MULTIPLICATION OF POLYNOMIALS.

• A Monomial times a multi-term polynomial. To do this, we have to expand the

brackets.

−2𝑥𝑥(𝑥𝑥2 + 3𝑥𝑥 − 4) = (−2𝑥𝑥) ∙ (𝑥𝑥2) + (−2𝑥𝑥) ∙ (3𝑥𝑥) + (−2𝑥𝑥) ∙ (−4) =

= −2𝑥𝑥3 − 6𝑥𝑥2 + 8𝑥𝑥


Unit 03 November

• A Multi-term polynomial times a multi-term polynomial. We have to multiply

every term by every term.

If 𝑃𝑃(𝑥𝑥) = 𝑥𝑥2 + 3𝑥𝑥 − 4 and 𝑅𝑅(𝑥𝑥) = −𝑥𝑥 + 3, find 𝑃𝑃(𝑥𝑥) ∙ 𝑅𝑅(𝑥𝑥)

𝑃𝑃(𝑥𝑥) ∙ 𝑅𝑅(𝑥𝑥) = (𝑥𝑥2 + 3𝑥𝑥 − 4 ) ∙ (−𝑥𝑥 + 3) =

= 𝑥𝑥2 ∙ (−𝑥𝑥 + 3) + 3𝑥𝑥 ∙ (−𝑥𝑥 + 3) − 4 ∙ (−𝑥𝑥 + 3) =

= −𝑥𝑥3 + 3𝑥𝑥2 − 3𝑥𝑥2 + 9𝑥𝑥 + 4𝑥𝑥 − 12 =

= −𝑥𝑥3 + 13𝑥𝑥 − 12

1.8. EXTRACTING FACTORS OF POLYNOMIALS.

To extract factors from polynomials we have to see with variables and factors

are repeating it.

Extract factors from 6𝑥𝑥2𝑦𝑦2 − 3𝑥𝑥𝑦𝑦2 + 30𝑥𝑥2𝑦𝑦

6𝑥𝑥2𝑦𝑦2 − 3𝑥𝑥𝑦𝑦2 + 30𝑥𝑥2𝑦𝑦 = (2𝑥𝑥𝑦𝑦) ∙ (3𝑥𝑥𝑦𝑦) + (−𝑦𝑦) ∙ (3𝑥𝑥𝑦𝑦) + (10𝑥𝑥) ∙ (3𝑥𝑥𝑦𝑦) =

= 3𝑥𝑥𝑦𝑦(2𝑥𝑥𝑦𝑦 − 𝑦𝑦 + 10𝑥𝑥)

2. POWER OF POLYNOMIALS.

The Power of a polynomial, 𝑷𝑷(𝒙𝒙)𝒏𝒏, is the multiplication of the polynomial

𝑷𝑷(𝒙𝒙), n times.

𝑷𝑷(𝒙𝒙)𝒏𝒏 = 𝑷𝑷(𝒙𝒙) ∙ 𝑷𝑷(𝒙𝒙) ∙ … ∙ 𝑷𝑷(𝒙𝒙)��𝒏𝒏 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕

2.1. BINOMIAL´S POWERS.

To solve the power of a binomial we have to use the Tartaglia's Triangle, also

known as Pascal´s Triangle.


Unit 03 November

To know the development of a binomial raised to the nth power we use the

nth+1 row of the triangle to find the 𝒄𝒄 coefficients.

(𝒂𝒂 + 𝒃𝒃)𝒏𝒏 = 𝟏𝟏𝒂𝒂𝒏𝒏𝒃𝒃𝟎𝟎 + 𝒄𝒄𝒏𝒏−𝟏𝟏𝒂𝒂𝒏𝒏−𝟏𝟏𝒃𝒃𝟏𝟏 + ⋯+ 𝒄𝒄𝟏𝟏𝒂𝒂𝟏𝟏𝒃𝒃𝒏𝒏−𝟏𝟏 + 𝟏𝟏𝒂𝒂𝟎𝟎𝒃𝒃𝒏𝒏

(𝒂𝒂 − 𝒃𝒃)𝒏𝒏 = �𝒂𝒂 + (−𝒃𝒃)�𝒏𝒏 = 𝟏𝟏𝒂𝒂𝒏𝒏(−𝒃𝒃)𝟎𝟎 + 𝒄𝒄𝒏𝒏−𝟏𝟏𝒂𝒂𝒏𝒏−𝟏𝟏(−𝒃𝒃)𝟏𝟏 + ⋯+ 𝒄𝒄𝟏𝟏𝒂𝒂𝟏𝟏(−𝒃𝒃)𝒏𝒏−𝟏𝟏 + 𝟏𝟏𝒂𝒂𝟎𝟎(−𝒃𝒃)𝒏𝒏

The exponents of 𝒂𝒂 and 𝒃𝒃 must add always 𝒏𝒏.

(𝑎𝑎 + 𝑏𝑏)2 ⇒ 𝑛𝑛 = 2 𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑙𝑙𝑓𝑓 𝑡𝑡ℎ𝑤𝑤 3𝑓𝑓𝑟𝑟 𝑓𝑓𝑙𝑙𝑤𝑤 ⇒ 𝑐𝑐𝑙𝑙𝑤𝑤𝑓𝑓𝑓𝑓𝑐𝑐𝑐𝑐𝑐𝑐𝑤𝑤𝑛𝑛𝑡𝑡𝑐𝑐 𝟏𝟏 𝟐𝟐 𝟏𝟏 ⇒

(𝑎𝑎 + 𝑏𝑏)2 = 𝟏𝟏𝑎𝑎2𝑏𝑏0 + 𝟐𝟐𝑎𝑎2−1𝑏𝑏0+1 + 𝟏𝟏𝑎𝑎2−2𝑏𝑏0+2 =

= 𝑎𝑎2 + 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2


Unit 03 November

(𝑎𝑎 − 𝑏𝑏)5 ⇒ 𝑛𝑛 = 5 𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑙𝑙𝑓𝑓 𝑡𝑡ℎ𝑤𝑤 6𝑡𝑡ℎ 𝑓𝑓𝑙𝑙𝑤𝑤 ⇒ 𝑐𝑐𝑙𝑙𝑤𝑤𝑓𝑓𝑓𝑓𝑐𝑐𝑐𝑐𝑐𝑐𝑤𝑤𝑛𝑛𝑡𝑡𝑐𝑐 𝟏𝟏 𝟓𝟓 𝟏𝟏𝟎𝟎 𝟏𝟏𝟎𝟎 𝟓𝟓 𝟏𝟏 ⇒

(𝑎𝑎 − 𝑏𝑏)5 = 𝟏𝟏𝑎𝑎5(−𝑏𝑏)0 + 𝟓𝟓𝑎𝑎5−1(−𝑏𝑏)0+1 + 𝟏𝟏𝟎𝟎𝑎𝑎5−2(−𝑏𝑏)0+2 + 𝟏𝟏𝟎𝟎𝑎𝑎5−3(−𝑏𝑏)0+3

+ 𝟓𝟓𝑎𝑎5−4(−𝑏𝑏)0+4 + 𝟏𝟏𝑎𝑎5−5(−𝑏𝑏)0+5 =

= 𝑎𝑎5 − 5𝑎𝑎4𝑏𝑏 + 10𝑎𝑎3𝑏𝑏2 − 10𝑎𝑎2𝑏𝑏3 + 5𝑎𝑎𝑏𝑏4 − 𝑏𝑏5

MATH VOCABULARY: Tartaglia's Triangle, Pascal's Triangle.

3. POLYNOMIAL IDENTITIES.

Some special products are called Polynomial Identities, and they serve to solve

some algebraic expressions. We will see three of them:

• The Square of the Sum: (𝒂𝒂 + 𝒃𝒃)𝟐𝟐 = 𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐

(2𝑥𝑥 + 𝑥𝑥2)2 = (2𝑥𝑥)2 + 2 ∙ (2𝑥𝑥) ∙ (𝑥𝑥2) + (𝑥𝑥2)2 = 4𝑥𝑥2 + 4𝑥𝑥3 + 𝑥𝑥4 =

= 𝑥𝑥4 + 4𝑥𝑥3 + 4𝑥𝑥2

• The Square of the Difference: (𝒂𝒂 − 𝒃𝒃)𝟐𝟐 = 𝒂𝒂𝟐𝟐 − 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐

(4𝑎𝑎 − 𝑏𝑏)2 = (4𝑎𝑎)2 − 2 ∙ (4𝑎𝑎) ∙ (𝑏𝑏) + (𝑏𝑏)2 = 16𝑎𝑎2 − 8𝑎𝑎𝑏𝑏 + 𝑏𝑏2

• The Product of a Sum and a Difference: (𝒂𝒂 + 𝒃𝒃) ∙ (𝒂𝒂 − 𝒃𝒃) = 𝒂𝒂𝟐𝟐 − 𝒃𝒃𝟐𝟐

(2𝑥𝑥 + 𝑦𝑦) ∙ (2𝑥𝑥 − 𝑦𝑦) = (2𝑥𝑥)2 − (𝑦𝑦)2 = 4𝑥𝑥2 − 𝑦𝑦2

MATH VOCABULARY: Polynomial Identities, Algebraic Expressions, Square of the Sum,

Square of the Difference, Product of a Sum and a Difference.


Unit 03 November

4. DIVISION OF POLYNOMIALS.

The division of polynomials is similar to the division of natural numbers. When

you divide polynomials you get a quotient and a remainder. In general, if you divide

the polynomial 𝑨𝑨(𝒙𝒙) by the polynomial 𝑩𝑩(𝒙𝒙) and the quotient and the remainder are

𝑸𝑸(𝒙𝒙) and 𝑹𝑹(𝒙𝒙) respectively.

⇒ 𝑨𝑨(𝒙𝒙) = 𝑩𝑩(𝒙𝒙) ∙ 𝑸𝑸(𝒙𝒙) + 𝑹𝑹(𝒙𝒙)

When the remainder is 𝟎𝟎, we have that 𝑨𝑨(𝒙𝒙) = 𝑩𝑩(𝒙𝒙) ∙ 𝑸𝑸(𝒙𝒙). In this case, the

polynomial 𝑨𝑨(𝒙𝒙) is divisible by 𝑩𝑩(𝒙𝒙), that is, 𝑩𝑩(𝒙𝒙) is a factor or divisor of 𝑨𝑨(𝒙𝒙).

Divide A(x) = 2𝑥𝑥5 − 7𝑥𝑥2 + 3𝑥𝑥 − 1 by 𝐵𝐵(𝑥𝑥) = 𝑥𝑥3 − 2𝑥𝑥2 + 1

+2𝑥𝑥5 −7𝑥𝑥2 +3𝑥𝑥 −1 𝑥𝑥3 −2𝑥𝑥2 +1

−2𝑥𝑥5 +4𝑥𝑥4 −2𝑥𝑥2 2𝑥𝑥2 +4𝑥𝑥 +8

+4𝑥𝑥4 −9𝑥𝑥2 +3𝑥𝑥 −1 𝑄𝑄(𝑥𝑥)

−4𝑥𝑥4 +8𝑥𝑥3 −4𝑥𝑥

+8𝑥𝑥3 −9𝑥𝑥2 −𝑥𝑥 −1

−8𝑥𝑥3 +16𝑥𝑥2 −8

+7𝑥𝑥2 −𝑥𝑥 −9 ⇐ 𝑅𝑅(𝑋𝑋)

⇒ 2𝑥𝑥5 − 7𝑥𝑥2 + 3𝑥𝑥 − 1 = (𝑥𝑥3 − 2𝑥𝑥2 + 1) ∙ (2𝑥𝑥2 + 4𝑥𝑥 + 8) + (7𝑥𝑥2 − 𝑥𝑥 − 9)


Unit 03 November

4.1. DIVISION OF A POLYNOMIAL BY (𝒙𝒙 − 𝒂𝒂). RUFFINI’S RULE.

It is very common to divide a polynomial by (𝐱𝐱 − 𝐚𝐚):

(3𝑥𝑥3 + 4𝑥𝑥 − 2) ÷ (𝑥𝑥 − 2) ⇒ 𝑄𝑄(𝑥𝑥) = 3𝑥𝑥2 + 6𝑥𝑥 + 16 𝑎𝑎𝑛𝑛𝑟𝑟 𝑅𝑅(𝑥𝑥) = 30

Using the above rules

But this division can also be done using Ruffini’s rule:

• Step 1: Set the coefficients of the dividend in one line. If the polynomial is not

complete, complete it by adding the missing terms with zeroes. Draw two

perpendicular lines like this:

3 0 4 −2

• Step 2: At the bottom left, place the opposite of the independent term of the

divisor:

3 0 4 −2

2

• Step 3: Bring down the first coefficient.

3 0 4 −2

2

3


Unit 03 November

• Step 4: Multiply this coefficient by the divisor and place it under the following

coefficient.

3 0 4 −2

2 6

3

• Step 5: Add the two coefficients.

3 0 4 −2

2 6

3 6

• Step 6: Repeat Steps 4 and 5 until you get the last number, like this:

3 0 4 −2

2 6 12 32

3 6 16 30

The last number obtained, 30, is the remainder of the division. The quotient is

a polynomial of one degree less than the dividend polynomial and whose coefficients

are the ones obtained in the division. The Coefficients of the Quotient are 3, 6 𝑎𝑎𝑛𝑛𝑟𝑟 16.

In this example, the quotient polynomial is:

𝑄𝑄(𝑥𝑥) = 3𝑥𝑥2 + 6𝑥𝑥 + 16

MATH VOCABULARY: Ruffini’s Rule.


Unit 03 November

4.2. RUFFINI’S RULE’S USES.

When the coefficients of a polynomial 𝑷𝑷(𝒙𝒙) are integers, if (𝒙𝒙 − 𝒂𝒂) is a factor

of 𝑷𝑷(𝒙𝒙) and “𝒂𝒂” is also an integer number, then “𝒂𝒂” is a divisor of the constant term of

𝑷𝑷(𝒙𝒙). So if you are looking for factors of a polynomial 𝑷𝑷(𝒙𝒙), have a try with the linear

factors (𝒙𝒙 − 𝒂𝒂) where “𝒂𝒂” is a divisor of the constant term of 𝑷𝑷(𝒙𝒙).

5. THE REMAINDER THEOREM.

Remember that you can calculate the number value of a polynomial at a given

value of the variable (1.5.). The Remainder Theorem states:

“The number value of the polynomial 𝑷𝑷(𝒙𝒙) at 𝒙𝒙 = 𝒂𝒂 is the same as the remainder of

the division 𝑷𝑷(𝒙𝒙) ÷ (𝒙𝒙 − 𝒂𝒂) . That is, 𝑷𝑷(𝒂𝒂) = 𝑹𝑹 .”

PROOF:

P(x) = (x − a) ∙ Q(x) + R

If x = a ⇒ P(a) = (a − a) ∙ Q(a) + R = 0 + R ⇒ P(a) = 𝑅𝑅

Find the remainder of this division using the Theorem:

(3𝑥𝑥3 + 2𝑥𝑥2 + 5𝑥𝑥 − 3) ÷ (𝑥𝑥 + 1)

Using the theorem:

𝑃𝑃(−1) = 3(−1)3 + 2(−1)2 + 5(−1) − 3 = −9 ⇒ 𝑅𝑅(𝑥𝑥) = −9

MATH VOCABULARY: Remainder Theorem.


Unit 03 November

6. ROOTS OF A POLYNOMIAL.

A number “𝒂𝒂” is called a Root of a polynomial 𝑷𝑷(𝒙𝒙) if 𝑷𝑷(𝒂𝒂) = 𝟎𝟎 . The roots (or

zeroes) of a polynomial are the solutions of the equation 𝑷𝑷(𝒙𝒙) = 𝟎𝟎.

One of the most important uses of Ruffini’s rule is to find the roots of a

polynomial.

Find the roots of 𝑃𝑃(𝑥𝑥) = 𝑥𝑥2 − 𝑥𝑥 − 2

The constant term is −2, so its divisors are ±1 𝑎𝑎𝑛𝑛𝑟𝑟 ± 2. Starting with 1:

1 −1 −2

1 1 0

1 0 −2 = 𝑅𝑅(𝑥𝑥) ≠ 0

1 −1 −2

−1 −1 +2

1 −2 0 = 𝑅𝑅(𝑥𝑥) ⇒ 𝑅𝑅𝑙𝑙𝑙𝑙𝑡𝑡

1 −1 −2

2 2 +2

1 1 0 = 𝑅𝑅(𝑥𝑥) ⇒ 𝑅𝑅𝑙𝑙𝑙𝑙𝑡𝑡

As the polynomial has a second degree we don´t need to test the last divisor. The roots

are −1 𝑎𝑎𝑛𝑛𝑟𝑟 2. We can also solve it using the remainder theorem

MATH VOCABULARY: Roots, Zeroes.


Unit 03 November

7. FACTORIZING POLYNOMIALS.

Factoring a polynomial means rewriting it as a product of polynomials of the

lowest degree as possible that can be multiplied together to give us the polynomial

that you started with.

𝑥𝑥2 − 16 = (𝑥𝑥 + 4) ∙ (𝑥𝑥 − 4)

There are different techniques for factorizing polynomials:

• Taking out common factor:

16𝑥𝑥3 − 2𝑥𝑥 = 2𝑥𝑥(8𝑥𝑥2 − 1)

• Using Polynomial Identities:

𝑥𝑥2 − 16 = (𝑥𝑥 + 4) ∙ (𝑥𝑥 − 4)

• Using the Fundamental Theorem of Algebra:

The roots of the polynomial 𝑃𝑃(𝑥𝑥) = 𝑥𝑥2 + 𝑥𝑥 − 6 are using the quadratic formula,

2 𝑎𝑎𝑛𝑛𝑟𝑟 − 3

So, you can rewrite: 𝑃𝑃(𝑥𝑥) = 𝑥𝑥2 + 𝑥𝑥 − 6 = (𝑥𝑥 − 2) ∙ (𝑥𝑥 + 3)

• Using Ruffini’s Rule:

𝑃𝑃(𝑥𝑥) = 𝑥𝑥3 − 2𝑥𝑥2 − 5𝑥𝑥 + 6


Unit 03 November

1 −2 −5 +6

1 1 −1 −6

1 −1 −6 0

−2 −2 +6

1 −3 0

⇒ 𝑃𝑃(𝑥𝑥) = 𝑥𝑥3 − 2𝑥𝑥2 − 5𝑥𝑥+ 6 = (𝑥𝑥 − 1) ∙ (𝑥𝑥+ 2) ∙ (𝑥𝑥 − 3)

• A combination of the previous ones:

𝑃𝑃(𝑥𝑥) = −2𝑥𝑥4 − 4𝑥𝑥3 + 14𝑥𝑥2 − 8𝑥𝑥

We can extract 𝑥𝑥, and 𝐻𝐻𝐻𝐻𝐻𝐻 (2,4,8,14) = 2

⇒ 𝑃𝑃(𝑥𝑥) = 2𝑥𝑥 ∙ (−𝑥𝑥3 − 2𝑥𝑥2 + 7𝑥𝑥 − 4) = 2𝑥𝑥 ∙ 𝑄𝑄(𝑥𝑥)

Factorizing 𝑄𝑄(𝑥𝑥), the divisors or 4: ± 1, ±2, ±4. We test with −1

−1 −2 +7 −4

−1 +1 +1 −8

−1 −1 +8 −12 = 𝑅𝑅(𝑥𝑥) ≠ 0 ⇒ 𝑁𝑁𝑙𝑙 𝑅𝑅𝑙𝑙𝑙𝑙𝑡𝑡

We test with +1:

−1 −2 +7 −4

+1 −1 −3 +4

−1 −3 +4 0 = 𝑅𝑅(𝑥𝑥) ⇒ 𝑅𝑅𝑙𝑙𝑙𝑙𝑡𝑡

⇒ (𝑥𝑥 − 1)𝑐𝑐𝑐𝑐 𝑎𝑎 𝑓𝑓𝑎𝑎𝑐𝑐𝑡𝑡𝑙𝑙𝑓𝑓 ⇒ 𝑃𝑃(𝑥𝑥) = 2𝑥𝑥 ∙ 𝑄𝑄(𝑥𝑥) = 2𝑥𝑥 ∙ (𝑥𝑥 − 1) ∙ (−𝑥𝑥2 − 3𝑥𝑥 + 4)

The last one can be factorizing using again Ruffini or using the quadratic formula:


Unit 03 November

𝑥𝑥 =−(−3) ± �(−3)2 − 4 ∙ (−1) ∙ 4

2 ∙ (−1) =3 ± 5−2 = �𝑥𝑥1 = −4

𝑥𝑥2 = +1

⇒ 𝑃𝑃(𝑥𝑥) = 2𝑥𝑥 ∙ (𝑥𝑥 − 1) ∙ (−𝑥𝑥2 − 3𝑥𝑥 + 4) = 2𝑥𝑥 ∙ (𝑥𝑥 − 1) ∙ (𝑥𝑥 − 1) ∙ (𝑥𝑥 + 4) =

= 2𝑥𝑥 ∙ (𝑥𝑥 − 1)2 ∙ (𝑥𝑥 + 4)

MATH VOCABULARY: Factorizing Polynomials, Fundamental Theorem of Algebra,

Quadratic Formula.

8. ALGEBRAIC FRACTIONS.

An Algebraic Fraction is the quotient of two polynomials, that is:

𝑷𝑷(𝒙𝒙)𝑸𝑸(𝒙𝒙)

2𝑥𝑥3 − 2𝑥𝑥 + 4𝑥𝑥 − 1 𝑐𝑐𝑐𝑐 𝑎𝑎𝑛𝑛 𝐴𝐴𝑙𝑙𝑙𝑙𝑤𝑤𝑏𝑏𝑓𝑓𝑎𝑎𝑐𝑐𝑐𝑐 𝐻𝐻𝑓𝑓𝑎𝑎𝑐𝑐𝑡𝑡𝑐𝑐𝑙𝑙𝑛𝑛

The same calculations that you do with numerical fractions can be done with

algebraic fractions. As you usually do with numerical fractions, you can simplify

algebraic fractions factoring the polynomials in the numerator and in the

denominator. Dividing by the H.C.F. of numerator and denominator you will get the

simplest form of the algebraic fraction.

𝑥𝑥6 − 6𝑥𝑥5 + 9𝑥𝑥4 + 4𝑥𝑥3 − 12𝑥𝑥2

𝑥𝑥3 − 2𝑥𝑥2 − 5𝑥𝑥 + 6 = 𝐻𝐻𝑎𝑎𝑐𝑐𝑡𝑡𝑙𝑙𝑓𝑓𝑐𝑐𝑧𝑧𝑐𝑐𝑛𝑛𝑙𝑙 =𝑥𝑥2 ∙ (𝑥𝑥 + 1) ∙ (𝑥𝑥 − 2)2 ∙ (𝑥𝑥 − 3)

(𝑥𝑥 − 1) ∙ (𝑥𝑥 + 2) ∙ (𝑥𝑥 − 3) =

=𝑥𝑥2 ∙ (𝑥𝑥 + 1) ∙ (𝑥𝑥 − 2)2

(𝑥𝑥 − 1) ∙ (𝑥𝑥 + 2) ∙(𝑥𝑥 − 3)(𝑥𝑥 − 3) =

𝑥𝑥2 ∙ (𝑥𝑥 + 1) ∙ (𝑥𝑥 − 2)2

(𝑥𝑥 − 1) ∙ (𝑥𝑥 + 2)


Unit 03 November

As you usually do with numerical fractions, you can also add, subtract, multiply

or divide algebraic fractions. (To add or subtract algebraic fractions you need to

reduce to common denominator).

MATH VOCABULARY: Algebraic Fraction.


4 eso academics - unit 03 - polynomials. algebraic fractions

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