4 ESO Academics - UNIT 03 - POLYNOMIALS. ALGEBRAIC FRACTIONS

1. 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).
2a
3
; 5𝑥𝑥; −2𝑦𝑦; 450𝑥𝑥2
𝑧𝑧 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 2𝑥𝑥2
𝑦𝑦3
𝑧𝑧 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

2. 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
(−2𝑥𝑥𝑦𝑦2
𝑧𝑧) ∙ (5𝑥𝑥2
𝑧𝑧3) = �(−2) ∙ 5� ∙ (𝑥𝑥1+2) ∙ (𝑦𝑦2+0) ∙ (𝑧𝑧1+3) = −10𝑥𝑥3
𝑦𝑦3
𝑧𝑧4
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
(12𝑎𝑎2
𝑏𝑏) ÷ (3𝑎𝑎) = (12 ÷ 3) ∙ (𝑎𝑎2−1) ∙ (𝑏𝑏1−0) = 4𝑎𝑎𝑎𝑎
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.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.2

3. 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.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.3

4. 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
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.4

5. 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 6𝑥𝑥2
𝑦𝑦2
− 3𝑥𝑥𝑦𝑦2
+ 30𝑥𝑥2
𝑦𝑦
6𝑥𝑥2
𝑦𝑦2
− 3𝑥𝑥𝑦𝑦2
+ 30𝑥𝑥2
𝑦𝑦 = (2𝑥𝑥𝑥𝑥) ∙ (3𝑥𝑥𝑥𝑥) + (−𝑦𝑦) ∙ (3𝑥𝑥𝑥𝑥) + (10𝑥𝑥) ∙ (3𝑥𝑥𝑥𝑥) =
= 3𝑥𝑥𝑥𝑥(2𝑥𝑥𝑥𝑥 − 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 𝒕𝒕𝒕𝒕 𝒕𝒕𝒕𝒕𝒕𝒕
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.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.5

6. 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
= 𝟏𝟏𝒂𝒂𝒏𝒏
𝒃𝒃𝟎𝟎
+ 𝒄𝒄𝒏𝒏−𝟏𝟏 𝒂𝒂𝒏𝒏−𝟏𝟏
𝒃𝒃𝟏𝟏
+ ⋯ + 𝒄𝒄𝟏𝟏 𝒂𝒂𝟏𝟏
𝒃𝒃𝒏𝒏−𝟏𝟏
+ 𝟏𝟏𝒂𝒂𝟎𝟎
𝒃𝒃𝒏𝒏
( 𝒂𝒂 − 𝒃𝒃)𝒏𝒏
= �𝒂𝒂 + (−𝒃𝒃)�
𝒏𝒏
= 𝟏𝟏𝒂𝒂𝒏𝒏(−𝒃𝒃)𝟎𝟎
+ 𝒄𝒄𝒏𝒏−𝟏𝟏 𝒂𝒂𝒏𝒏−𝟏𝟏(−𝒃𝒃)𝟏𝟏
+ ⋯ + 𝒄𝒄𝟏𝟏 𝒂𝒂𝟏𝟏(−𝒃𝒃)𝒏𝒏−𝟏𝟏
+ 𝟏𝟏𝒂𝒂𝟎𝟎(−𝒃𝒃)𝒏𝒏
The exponents of 𝒂𝒂 and 𝒃𝒃 must add always 𝒏𝒏.
(𝑎𝑎 + 𝑏𝑏)2
⇒ 𝑛𝑛 = 2 𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 3𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 ⇒ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟏𝟏 𝟐𝟐 𝟏𝟏 ⇒
(𝑎𝑎 + 𝑏𝑏)2
= 𝟏𝟏𝑎𝑎2
𝑏𝑏0
+ 𝟐𝟐𝑎𝑎2−1
𝑏𝑏0+1
+ 𝟏𝟏𝑎𝑎2−2
𝑏𝑏0+2
=
= 𝑎𝑎2
+ 2𝑎𝑎𝑎𝑎 + 𝑏𝑏2
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.6

7. Unit 03 November
(𝑎𝑎 − 𝑏𝑏)5
⇒ 𝑛𝑛 = 5 𝑤𝑤𝑤𝑤 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 6𝑡𝑡ℎ 𝑟𝑟𝑟𝑟𝑟𝑟 ⇒ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝟏𝟏 𝟓𝟓 𝟏𝟏𝟏𝟏 𝟏𝟏𝟏𝟏 𝟓𝟓 𝟏𝟏 ⇒
(𝑎𝑎 − 𝑏𝑏)5
= 𝟏𝟏𝑎𝑎5(−𝑏𝑏)0
+ 𝟓𝟓𝑎𝑎5−1(−𝑏𝑏)0+1
+ 𝟏𝟏𝟏𝟏𝑎𝑎5−2(−𝑏𝑏)0+2
+ 𝟏𝟏𝟏𝟏𝑎𝑎5−3(−𝑏𝑏)0+3
+ 𝟓𝟓𝑎𝑎5−4(−𝑏𝑏)0+4
+ 𝟏𝟏𝑎𝑎5−5(−𝑏𝑏)0+5
=
= 𝑎𝑎5
− 5𝑎𝑎4
𝑏𝑏 + 10𝑎𝑎3
𝑏𝑏2
− 10𝑎𝑎2
𝑏𝑏3
+ 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
+ 𝟐𝟐𝟐𝟐𝟐𝟐 + 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
− 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒃𝒃𝟐𝟐
(4𝑎𝑎 − 𝑏𝑏)2
= (4𝑎𝑎)2
− 2 ∙ (4𝑎𝑎) ∙ (𝑏𝑏) + (𝑏𝑏)2
= 16𝑎𝑎2
− 8𝑎𝑎𝑎𝑎 + 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.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.7

8. 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)
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.8

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 𝑎𝑎𝑎𝑎𝑎𝑎 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
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.9

10. 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 𝑎𝑎𝑎𝑎𝑎𝑎 16.
In this example, the quotient polynomial is:
𝑄𝑄(𝑥𝑥) = 3𝑥𝑥2
+ 6𝑥𝑥 + 16
MATH VOCABULARY: Ruffini’s Rule.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.10

11. 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.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.11

12. 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 𝑎𝑎𝑎𝑎𝑎𝑎 ± 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) ⇒ 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
1 −1 −2
2 2 +2
1 1 0 = 𝑅𝑅(𝑥𝑥) ⇒ 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
As the polynomial has a second degree we don´t need to test the last divisor. The roots
are −1 𝑎𝑎𝑎𝑎𝑎𝑎 2. We can also solve it using the remainder theorem
MATH VOCABULARY: Roots, Zeroes.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.12

13. 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 𝑎𝑎𝑎𝑎𝑎𝑎 − 3
So, you can rewrite: 𝑃𝑃(𝑥𝑥) = 𝑥𝑥2
+ 𝑥𝑥 − 6 = (𝑥𝑥 − 2) ∙ (𝑥𝑥 + 3)
• Using Ruffini’s Rule:
𝑃𝑃(𝑥𝑥) = 𝑥𝑥3
− 2𝑥𝑥2
− 5𝑥𝑥 + 6
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.13

14. 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 ⇒ 𝑁𝑁𝑁𝑁 𝑅𝑅𝑅𝑅𝑅𝑅𝑡𝑡
We test with +1:
−1 −2 +7 −4
+1 −1 −3 +4
−1 −3 +4 0 = 𝑅𝑅(𝑥𝑥) ⇒ 𝑅𝑅𝑅𝑅𝑅𝑅𝑡𝑡
⇒ (𝑥𝑥 − 1)𝑖𝑖𝑖𝑖 𝑎𝑎 𝑓𝑓𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ⇒ 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:
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.14

15. 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 + 4
𝑥𝑥 − 1
𝑖𝑖𝑠𝑠 𝑎𝑎𝑛𝑛 𝐴𝐴𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐹𝐹𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
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
= 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 =
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)
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.15

16. 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.
Axel Cotón Gutiérrez Mathematics 4º ESO 4.3.16