4 eso academics - unit 03 - polynomials. algebraic fractions
TRANSCRIPT
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
; 5π₯π₯; β2π¦π¦; 450π₯π₯2π§π§ 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 β5π₯π₯3 is 3
The degree of 2π₯π₯2π¦π¦3π§π§ 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 CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.1
Unit 03 November
3π₯π₯ + 5π₯π₯ = 8π₯π₯
3π₯π₯ β 2π₯π₯2 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.
2π₯π₯7 β 3π₯π₯3 = (2 β 3) β π₯π₯7+3 = 6π₯π₯10
(β2π₯π₯π¦π¦2π§π§) β (5π₯π₯2π§π§3) = οΏ½(β2) β 5οΏ½ β (π₯π₯1+2) β (π¦π¦2+0) β (π§π§1+3) = β10π₯π₯3π¦π¦3π§π§4
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.
10π₯π₯5 Γ· 2π₯π₯2 = (10 Γ· 2) β (π₯π₯5β2) = 5π₯π₯3
(12ππ2ππ) Γ· (3ππ) = (12 Γ· 3) β (ππ2β1) β (ππ1β0) = 4ππππ
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 CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.2
Unit 03 November
You usually write polynomials with the terms in βDecreasingβ 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 CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.3
Unit 03 November
1.5. EVALUATING POLYNOMIALS.
βEvaluatingβ a polynomial π·π·(ππ) is calculating its numerical value at a given
value of the variable: ππ = ππ. You must substitute the variable ππ for the value ππ, and
calculate the value of the polynomial π·π·(ππ).
Evaluate ππ(π₯π₯) = π₯π₯4 β 3π₯π₯2 + π₯π₯ + 1 at π₯π₯ = 2
ππ(2) = 24 β 3 β 22 + 2 + 1 = 16 β 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 ππ(π₯π₯) = π₯π₯2 + 3π₯π₯ β 4;ππ(π₯π₯) = π₯π₯3 + 2π₯π₯ + 1 and π π (π₯π₯) = βπ₯π₯ + 3
Find ππ(π₯π₯) + ππ(π₯π₯)β π π (π₯π₯) = ππ(π₯π₯)
ππ(π₯π₯) = (π₯π₯2 + 3π₯π₯ β 4) + (π₯π₯3 + 2π₯π₯ + 1)β (βπ₯π₯ + 3) =
= π₯π₯3 + π₯π₯2 + οΏ½3π₯π₯ + 2π₯π₯ β (βπ₯π₯)οΏ½ + οΏ½(β4) + 1 β 3οΏ½ = π₯π₯3 + π₯π₯2 + 6π₯π₯ β 6
1.7. MULTIPLICATION OF POLYNOMIALS.
β’ A Monomial times a multi-term polynomial. To do this, we have to expand the
brackets.
β2π₯π₯(π₯π₯2 + 3π₯π₯ β 4) = (β2π₯π₯) β (π₯π₯2) + (β2π₯π₯) β (3π₯π₯) + (β2π₯π₯) β (β4) =
= β2π₯π₯3 β 6π₯π₯2 + 8π₯π₯
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.4
Unit 03 November
β’ A Multi-term polynomial times a multi-term polynomial. We have to multiply
every term by every term.
If ππ(π₯π₯) = π₯π₯2 + 3π₯π₯ β 4 and π π (π₯π₯) = βπ₯π₯ + 3, find ππ(π₯π₯) β π π (π₯π₯)
ππ(π₯π₯) β π π (π₯π₯) = (π₯π₯2 + 3π₯π₯ β 4 ) β (βπ₯π₯ + 3) =
= π₯π₯2 β (βπ₯π₯ + 3) + 3π₯π₯ β (βπ₯π₯ + 3) β 4 β (βπ₯π₯ + 3) =
= βπ₯π₯3 + 3π₯π₯2 β 3π₯π₯2 + 9π₯π₯ + 4π₯π₯ β 12 =
= βπ₯π₯3 + 13π₯π₯ β 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 6π₯π₯2π¦π¦2 β 3π₯π₯π¦π¦2 + 30π₯π₯2π¦π¦
6π₯π₯2π¦π¦2 β 3π₯π₯π¦π¦2 + 30π₯π₯2π¦π¦ = (2π₯π₯π¦π¦) β (3π₯π₯π¦π¦) + (βπ¦π¦) β (3π₯π₯π¦π¦) + (10π₯π₯) β (3π₯π₯π¦π¦) =
= 3π₯π₯π¦π¦(2π₯π₯π¦π¦ β π¦π¦ + 10π₯π₯)
2. POWER OF POLYNOMIALS.
The Power of a polynomial, π·π·(ππ)ππ, is the multiplication of the polynomial
π·π·(ππ), n times.
π·π·(ππ)ππ = π·π·(ππ) β π·π·(ππ) β β¦ β π·π·(ππ)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ ππππππππππ
2.1. BINOMIALΒ΄S POWERS.
To solve the power of a binomial we have to use the Tartaglia's Triangle, also
known as PascalΒ΄s Triangle.
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.5
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 ππ coefficients.
(ππ + ππ)ππ = ππππππππππ + ππππβππππππβππππππ + β―+ ππππππππππππβππ + ππππππππππ
(ππ β ππ)ππ = οΏ½ππ + (βππ)οΏ½ππ = ππππππ(βππ)ππ + ππππβππππππβππ(βππ)ππ + β―+ ππππππππ(βππ)ππβππ + ππππππ(βππ)ππ
The exponents of ππ and ππ must add always ππ.
(ππ + ππ)2 β ππ = 2 π€π€π€π€ ππππππππ ππππππ π‘π‘βπ€π€ 3ππππ πππππ€π€ β πππππ€π€πππππππππππ€π€πππ‘π‘ππ ππ ππ ππ β
(ππ + ππ)2 = ππππ2ππ0 + ππππ2β1ππ0+1 + ππππ2β2ππ0+2 =
= ππ2 + 2ππππ + ππ2
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.6
Unit 03 November
(ππ β ππ)5 β ππ = 5 π€π€π€π€ ππππππππ ππππππ π‘π‘βπ€π€ 6π‘π‘β πππππ€π€ β πππππ€π€πππππππππππ€π€πππ‘π‘ππ ππ ππ ππππ ππππ ππ ππ β
(ππ β ππ)5 = ππππ5(βππ)0 + ππππ5β1(βππ)0+1 + ππππππ5β2(βππ)0+2 + ππππππ5β3(βππ)0+3
+ ππππ5β4(βππ)0+4 + ππππ5β5(βππ)0+5 =
= ππ5 β 5ππ4ππ + 10ππ3ππ2 β 10ππ2ππ3 + 5ππππ4 β ππ5
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:
β’ The Square of the Sum: (ππ + ππ)ππ = ππππ + ππππππ + ππππ
(2π₯π₯ + π₯π₯2)2 = (2π₯π₯)2 + 2 β (2π₯π₯) β (π₯π₯2) + (π₯π₯2)2 = 4π₯π₯2 + 4π₯π₯3 + π₯π₯4 =
= π₯π₯4 + 4π₯π₯3 + 4π₯π₯2
β’ The Square of the Difference: (ππ β ππ)ππ = ππππ β ππππππ + ππππ
(4ππ β ππ)2 = (4ππ)2 β 2 β (4ππ) β (ππ) + (ππ)2 = 16ππ2 β 8ππππ + ππ2
β’ The Product of a Sum and a Difference: (ππ + ππ) β (ππ β ππ) = ππππ β ππππ
(2π₯π₯ + π¦π¦) β (2π₯π₯ β π¦π¦) = (2π₯π₯)2 β (π¦π¦)2 = 4π₯π₯2 β π¦π¦2
MATH VOCABULARY: Polynomial Identities, Algebraic Expressions, Square of the Sum,
Square of the Difference, Product of a Sum and a Difference.
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.7
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 π¨π¨(ππ) by the polynomial π©π©(ππ) and the quotient and the remainder are
πΈπΈ(ππ) and πΉπΉ(ππ) respectively.
β π¨π¨(ππ) = π©π©(ππ) β πΈπΈ(ππ) + πΉπΉ(ππ)
When the remainder is ππ, we have that π¨π¨(ππ) = π©π©(ππ) β πΈπΈ(ππ). In this case, the
polynomial π¨π¨(ππ) is divisible by π©π©(ππ), that is, π©π©(ππ) is a factor or divisor of π¨π¨(ππ).
Divide A(x) = 2π₯π₯5 β 7π₯π₯2 + 3π₯π₯ β 1 by π΅π΅(π₯π₯) = π₯π₯3 β 2π₯π₯2 + 1
+2π₯π₯5 β7π₯π₯2 +3π₯π₯ β1 π₯π₯3 β2π₯π₯2 +1
β2π₯π₯5 +4π₯π₯4 β2π₯π₯2 2π₯π₯2 +4π₯π₯ +8
+4π₯π₯4 β9π₯π₯2 +3π₯π₯ β1 ππ(π₯π₯)
β4π₯π₯4 +8π₯π₯3 β4π₯π₯
+8π₯π₯3 β9π₯π₯2 βπ₯π₯ β1
β8π₯π₯3 +16π₯π₯2 β8
+7π₯π₯2 βπ₯π₯ β9 β π π (ππ)
β 2π₯π₯5 β 7π₯π₯2 + 3π₯π₯ β 1 = (π₯π₯3 β 2π₯π₯2 + 1) β (2π₯π₯2 + 4π₯π₯ + 8) + (7π₯π₯2 β π₯π₯ β 9)
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.8
Unit 03 November
4.1. DIVISION OF A POLYNOMIAL BY (ππ β ππ). RUFFINIβS RULE.
It is very common to divide a polynomial by (π±π± β ππ):
(3π₯π₯3 + 4π₯π₯ β 2) Γ· (π₯π₯ β 2) β ππ(π₯π₯) = 3π₯π₯2 + 6π₯π₯ + 16 ππππππ π π (π₯π₯) = 30
Using the above rules
But this division can also be done using Ruffiniβs rule:
β’ 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 β2
β’ Step 2: At the bottom left, place the opposite of the independent term of the
divisor:
3 0 4 β2
2
β’ Step 3: Bring down the first coefficient.
3 0 4 β2
2
3
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.9
Unit 03 November
β’ Step 4: Multiply this coefficient by the divisor and place it under the following
coefficient.
3 0 4 β2
2 6
3
β’ Step 5: Add the two coefficients.
3 0 4 β2
2 6
3 6
β’ Step 6: Repeat Steps 4 and 5 until you get the last number, like this:
3 0 4 β2
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 ππππππ 16.
In this example, the quotient polynomial is:
ππ(π₯π₯) = 3π₯π₯2 + 6π₯π₯ + 16
MATH VOCABULARY: Ruffiniβs Rule.
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.10
Unit 03 November
4.2. RUFFINIβS RULEβS USES.
When the coefficients of a polynomial π·π·(ππ) are integers, if (ππ β ππ) is a factor
of π·π·(ππ) and βππβ is also an integer number, then βππβ is a divisor of the constant term of
π·π·(ππ). So if you are looking for factors of a polynomial π·π·(ππ), have a try with the linear
factors (ππ β ππ) where βππβ is a divisor of the constant term of π·π·(ππ).
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:
βThe number value of the polynomial π·π·(ππ) at ππ = ππ is the same as the remainder of
the division π·π·(ππ) Γ· (ππ β ππ) . That is, π·π·(ππ) = πΉπΉ .β
PROOF:
P(x) = (x β a) β Q(x) + R
If x = a β P(a) = (a β a) β Q(a) + R = 0 + R β P(a) = π π
Find the remainder of this division using the Theorem:
(3π₯π₯3 + 2π₯π₯2 + 5π₯π₯ β 3) Γ· (π₯π₯ + 1)
Using the theorem:
ππ(β1) = 3(β1)3 + 2(β1)2 + 5(β1) β 3 = β9 β π π (π₯π₯) = β9
MATH VOCABULARY: Remainder Theorem.
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.11
Unit 03 November
6. ROOTS OF A POLYNOMIAL.
A number βππβ is called a Root of a polynomial π·π·(ππ) if π·π·(ππ) = ππ . The roots (or
zeroes) of a polynomial are the solutions of the equation π·π·(ππ) = ππ.
One of the most important uses of Ruffiniβs rule is to find the roots of a
polynomial.
Find the roots of ππ(π₯π₯) = π₯π₯2 β π₯π₯ β 2
The constant term is β2, so its divisors are Β±1 ππππππ Β± 2. Starting with 1:
1 β1 β2
1 1 0
1 0 β2 = π π (π₯π₯) β 0
1 β1 β2
β1 β1 +2
1 β2 0 = π π (π₯π₯) β π π πππππ‘π‘
1 β1 β2
2 2 +2
1 1 0 = π π (π₯π₯) β π π πππππ‘π‘
As the polynomial has a second degree we donΒ΄t need to test the last divisor. The roots
are β1 ππππππ 2. We can also solve it using the remainder theorem
MATH VOCABULARY: Roots, Zeroes.
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.12
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.
π₯π₯2 β 16 = (π₯π₯ + 4) β (π₯π₯ β 4)
There are different techniques for factorizing polynomials:
β’ Taking out common factor:
16π₯π₯3 β 2π₯π₯ = 2π₯π₯(8π₯π₯2 β 1)
β’ Using Polynomial Identities:
π₯π₯2 β 16 = (π₯π₯ + 4) β (π₯π₯ β 4)
β’ Using the Fundamental Theorem of Algebra:
The roots of the polynomial ππ(π₯π₯) = π₯π₯2 + π₯π₯ β 6 are using the quadratic formula,
2 ππππππ β 3
So, you can rewrite: ππ(π₯π₯) = π₯π₯2 + π₯π₯ β 6 = (π₯π₯ β 2) β (π₯π₯ + 3)
β’ Using Ruffiniβs Rule:
ππ(π₯π₯) = π₯π₯3 β 2π₯π₯2 β 5π₯π₯ + 6
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.13
Unit 03 November
1 β2 β5 +6
1 1 β1 β6
1 β1 β6 0
β2 β2 +6
1 β3 0
β ππ(π₯π₯) = π₯π₯3 β 2π₯π₯2 β 5π₯π₯+ 6 = (π₯π₯ β 1) β (π₯π₯+ 2) β (π₯π₯ β 3)
β’ A combination of the previous ones:
ππ(π₯π₯) = β2π₯π₯4 β 4π₯π₯3 + 14π₯π₯2 β 8π₯π₯
We can extract π₯π₯, and π»π»π»π»π»π» (2,4,8,14) = 2
β ππ(π₯π₯) = 2π₯π₯ β (βπ₯π₯3 β 2π₯π₯2 + 7π₯π₯ β 4) = 2π₯π₯ β ππ(π₯π₯)
Factorizing ππ(π₯π₯), the divisors or 4: Β± 1, Β±2, Β±4. We test with β1
β1 β2 +7 β4
β1 +1 +1 β8
β1 β1 +8 β12 = π π (π₯π₯) β 0 β ππππ π π πππππ‘π‘
We test with +1:
β1 β2 +7 β4
+1 β1 β3 +4
β1 β3 +4 0 = π π (π₯π₯) β π π πππππ‘π‘
β (π₯π₯ β 1)ππππ ππ πππππππ‘π‘ππππ β ππ(π₯π₯) = 2π₯π₯ β ππ(π₯π₯) = 2π₯π₯ β (π₯π₯ β 1) β (βπ₯π₯2 β 3π₯π₯ + 4)
The last one can be factorizing using again Ruffini or using the quadratic formula:
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.14
Unit 03 November
π₯π₯ =β(β3) Β± οΏ½(β3)2 β 4 β (β1) β 4
2 β (β1) =3 Β± 5β2 = οΏ½π₯π₯1 = β4
π₯π₯2 = +1
β ππ(π₯π₯) = 2π₯π₯ β (π₯π₯ β 1) β (βπ₯π₯2 β 3π₯π₯ + 4) = 2π₯π₯ β (π₯π₯ β 1) β (π₯π₯ β 1) β (π₯π₯ + 4) =
= 2π₯π₯ β (π₯π₯ β 1)2 β (π₯π₯ + 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:
π·π·(ππ)πΈπΈ(ππ)
2π₯π₯3 β 2π₯π₯ + 4π₯π₯ β 1 ππππ ππππ π΄π΄πππππ€π€ππππππππππ π»π»πππππππ‘π‘ππππππ
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.
π₯π₯6 β 6π₯π₯5 + 9π₯π₯4 + 4π₯π₯3 β 12π₯π₯2
π₯π₯3 β 2π₯π₯2 β 5π₯π₯ + 6 = π»π»πππππ‘π‘πππππππ§π§ππππππ =π₯π₯2 β (π₯π₯ + 1) β (π₯π₯ β 2)2 β (π₯π₯ β 3)
(π₯π₯ β 1) β (π₯π₯ + 2) β (π₯π₯ β 3) =
=π₯π₯2 β (π₯π₯ + 1) β (π₯π₯ β 2)2
(π₯π₯ β 1) β (π₯π₯ + 2) β(π₯π₯ β 3)(π₯π₯ β 3) =
π₯π₯2 β (π₯π₯ + 1) β (π₯π₯ β 2)2
(π₯π₯ β 1) β (π₯π₯ + 2)
Axel CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.15
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 CotΓ³n GutiΓ©rrez Mathematics 4ΒΊ ESO 4.3.16