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FACTORING
CASES
B Y A N D R É S A L F O N S O P O L O
MAIN CASES OF FACTORIZATION
1) common factor
2) common factor by grouping of terms
3) Difference of Perfect Squares
4) Perfect Square Trinomial
5) Trinomial of the form 𝒙𝟐 + 𝒃𝒙 + 𝒄
6) trinomial of the form 𝐚𝒙𝟐
+ 𝒃𝒙 + 𝒄
7) Sum and Difference of Perfect
Cubes
1. COMMON FACTOR
 3𝑥 + 3𝑦
 𝑚𝑝 + 𝑚𝑞 + 𝑚𝑟
 𝑎6 − 3𝑎4 + 8𝑎3 − 4𝑎2
 2𝑥9
Applies to binomials,
trinomials and polynomials
of four terms or more.
It is the first case to be to
inspect when trying to factor
a polynomial.
 The common factor can be a
number, a letter letter, several
letters, a negative sign, an
algebraic expression (enclosed in
parentheses) or combinations of
all of the above
1. COMMON FACTOR
HOW TO FACTOR?
 From the coefficients of the terms, the GCD
(Greatest Common Divisor) of the terms is
extracted.
 From the repeated letters or expressions in
parentheses, the one with the lowest exponent is
extracted.
 The common factor is written, followed by a
parenthesis where the polynomial that remains
after the common factor has left each term is
written.
EXERCISES
 3𝑥 + 3𝑦
 𝑚𝑝 + 𝑚𝑞 + 𝑚𝑟
 𝑎6
− 3𝑎4
+ 8𝑎3
− 4𝑎2
2. COMMON FACTOR BY
GROUPING OF TERMS
 It is applied in polynomials that have 4,
6, 8 or more terms (as long as the
number is even) and where it has already
been verified that there is no common
factor (case 1).
2. COMMON FACTOR BY
GROUPING OF TERMS
HOW TO FACTOR?
EXERCISES
 𝑝𝑥 + 𝑚𝑥 + 𝑝𝑦 + 𝑚𝑦
 2𝑎𝑐 − 5𝑏𝑑 − 2𝑎 + 2𝑎𝑑 + 5b-
5bc
 We form groups, generally with the
same number of terms that have
the same common factor.
 We factor in each group as in the
previous case
 There must now be a new common
factor between the terms, so we
factor again as in the first case.
3. DIFFERENCE OF PERFECT SQUARES
It applies only to binomials, where the
first term is positive and the second
term is negative.
It is recognized because the
coefficients of the terms are perfect
square numbers
 𝑎2 − 𝑏2
 4𝑎2
− 9
3. DIFFERENCE OF PERFECT SQUARES
HOW TO FACTOR? EXERCISES
 The square root of each term is extracted.
 - Two groups of parentheses (connected to
each other by multiplication) are opened.
 - The square roots that were obtained from
each term are noted inside each parenthesis
 𝑎2 − 𝑏2
 4𝑎2
− 9
 16𝑥2
− 25𝑦4
4.PERFECT SQUARE TRINOMIAL
 - The trinomial must be organized in ascending
or descending order (either of the two).
 - Both the first and third terms must be
positive.
 Likewise, those two terms must be perfect
squares (that is, they must have an exact
square root).
 4𝑥2
+ 12𝑥𝑦2
+ 9𝑦4
 25𝑚4 − 40𝑚2 + 16
4.PERFECT SQUARE TRINOMIAL
HOW TO FACTOR?
• we extract the square root of the first and
third terms
• write them in parentheses, separating them by
the sign that accompanies the second term;
• when closing the parenthesis we raise the
whole binomial to the square.
EXERCISES
 4𝑥2 + 12𝑥𝑦2 + 9𝑦4
 25𝑚4 − 40𝑚2 + 16
5. TRINOMIAL OF THE FORM 𝒙𝟐
+ 𝒃𝒙 + 𝒄
 The trinomial must be organized in descending
order.
 The coefficient of the first term must be one
(1).
 The degree (exponent) of the first term must
be twice the degree (exponent) of the second
term.
 𝑥2 + 5𝑥 + 6
 𝑎2 − 2𝑎 − 15
Example
5. TRINOMIAL OF THE FORM 𝒙𝟐
+ 𝒃𝒙 + 𝒄
 Two groups of parentheses are opened.
 The square root of the first term is
extracted and noted at the beginning of
each parenthesis.
 The signs are defined:
 We are looking for two quantities that
multiplied give the independent term as a
result (ie c), and that added give the
coefficient of the second term as a result
(ie b).
 we write the quantities in the blanks
EXERCISES
HOW TO FACTOR?
 𝑥2 − 2𝑥 − 15
 𝑥4
+ 11𝑥2
+ 28
6.TRINOMIAL OF THE FORM 𝐚𝒙𝟐
+ 𝒃𝒙 + 𝒄
 The trinomial must be organized in
descending order.
 The first term must be positive and different
from one (a≠1).
 The degree (exponent) of the first term must
be twice the degree (exponent) of the second
term.
EXAMPLE
• 2𝑥2 + 11𝑥 + 5
• 7𝑚2
− 23𝑚 + 6
6.TRINOMIAL OF THE FORM 𝐚𝒙𝟐
+ 𝒃𝒙 + 𝒄
HOW TO FACTOR?
 multiply and divide the trinomial by the main
coefficient, that is, a.
 In the numerator we carry out the distributive
property, in the second term the product is not
realized but is left expressed
 - The first term is expressed as the square of what
was left in parentheses in the second term.
 - We apply case 5 (Trinomial of the form
x2n+bxn+c) in the numerator.
 - We apply case 1 (Common Factor) in the
parentheses formed and we simplify the fraction
(to eliminate the denominator).
EXERCISE
6𝑥2
+ 5𝑥 − 4
7. SUM AND DIFFERENCE OF PERFECT
CUBES
It applies only in binomials, where the
first term is positive (the second term
can be positive or negative).
- It is recognized because the
coefficients of the terms are perfect
cube numbers
7. SUM AND DIFFERENCE OF PERFECT
CUBES
HOW TO FACTOR?
• two parentheses are opened
• In the first parenthesis, put the cube roots of
each of the terms and the sign that
accompanies them.
• In the second parenthesis we write the first
term squared, minus (plus) the product of the
first by the second, plus the square of the third
term.
EXERCISES
8𝑥3 − 27𝑦3
27𝑥3 + 125𝑦9
Factoring cases - Andrés.pptx

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Factoring cases - Andrés.pptx

  • 1. FACTORING CASES B Y A N D R É S A L F O N S O P O L O
  • 2. MAIN CASES OF FACTORIZATION 1) common factor 2) common factor by grouping of terms 3) Difference of Perfect Squares 4) Perfect Square Trinomial 5) Trinomial of the form 𝒙𝟐 + 𝒃𝒙 + 𝒄 6) trinomial of the form 𝐚𝒙𝟐 + 𝒃𝒙 + 𝒄 7) Sum and Difference of Perfect Cubes
  • 3. 1. COMMON FACTOR  3𝑥 + 3𝑦  𝑚𝑝 + 𝑚𝑞 + 𝑚𝑟  𝑎6 − 3𝑎4 + 8𝑎3 − 4𝑎2  2𝑥9 Applies to binomials, trinomials and polynomials of four terms or more. It is the first case to be to inspect when trying to factor a polynomial.  The common factor can be a number, a letter letter, several letters, a negative sign, an algebraic expression (enclosed in parentheses) or combinations of all of the above
  • 4. 1. COMMON FACTOR HOW TO FACTOR?  From the coefficients of the terms, the GCD (Greatest Common Divisor) of the terms is extracted.  From the repeated letters or expressions in parentheses, the one with the lowest exponent is extracted.  The common factor is written, followed by a parenthesis where the polynomial that remains after the common factor has left each term is written. EXERCISES  3𝑥 + 3𝑦  𝑚𝑝 + 𝑚𝑞 + 𝑚𝑟  𝑎6 − 3𝑎4 + 8𝑎3 − 4𝑎2
  • 5. 2. COMMON FACTOR BY GROUPING OF TERMS  It is applied in polynomials that have 4, 6, 8 or more terms (as long as the number is even) and where it has already been verified that there is no common factor (case 1).
  • 6. 2. COMMON FACTOR BY GROUPING OF TERMS HOW TO FACTOR? EXERCISES  𝑝𝑥 + 𝑚𝑥 + 𝑝𝑦 + 𝑚𝑦  2𝑎𝑐 − 5𝑏𝑑 − 2𝑎 + 2𝑎𝑑 + 5b- 5bc  We form groups, generally with the same number of terms that have the same common factor.  We factor in each group as in the previous case  There must now be a new common factor between the terms, so we factor again as in the first case.
  • 7. 3. DIFFERENCE OF PERFECT SQUARES It applies only to binomials, where the first term is positive and the second term is negative. It is recognized because the coefficients of the terms are perfect square numbers  𝑎2 − 𝑏2  4𝑎2 − 9
  • 8. 3. DIFFERENCE OF PERFECT SQUARES HOW TO FACTOR? EXERCISES  The square root of each term is extracted.  - Two groups of parentheses (connected to each other by multiplication) are opened.  - The square roots that were obtained from each term are noted inside each parenthesis  𝑎2 − 𝑏2  4𝑎2 − 9  16𝑥2 − 25𝑦4
  • 9. 4.PERFECT SQUARE TRINOMIAL  - The trinomial must be organized in ascending or descending order (either of the two).  - Both the first and third terms must be positive.  Likewise, those two terms must be perfect squares (that is, they must have an exact square root).  4𝑥2 + 12𝑥𝑦2 + 9𝑦4  25𝑚4 − 40𝑚2 + 16
  • 10. 4.PERFECT SQUARE TRINOMIAL HOW TO FACTOR? • we extract the square root of the first and third terms • write them in parentheses, separating them by the sign that accompanies the second term; • when closing the parenthesis we raise the whole binomial to the square. EXERCISES  4𝑥2 + 12𝑥𝑦2 + 9𝑦4  25𝑚4 − 40𝑚2 + 16
  • 11. 5. TRINOMIAL OF THE FORM 𝒙𝟐 + 𝒃𝒙 + 𝒄  The trinomial must be organized in descending order.  The coefficient of the first term must be one (1).  The degree (exponent) of the first term must be twice the degree (exponent) of the second term.  𝑥2 + 5𝑥 + 6  𝑎2 − 2𝑎 − 15 Example
  • 12. 5. TRINOMIAL OF THE FORM 𝒙𝟐 + 𝒃𝒙 + 𝒄  Two groups of parentheses are opened.  The square root of the first term is extracted and noted at the beginning of each parenthesis.  The signs are defined:  We are looking for two quantities that multiplied give the independent term as a result (ie c), and that added give the coefficient of the second term as a result (ie b).  we write the quantities in the blanks EXERCISES HOW TO FACTOR?  𝑥2 − 2𝑥 − 15  𝑥4 + 11𝑥2 + 28
  • 13. 6.TRINOMIAL OF THE FORM 𝐚𝒙𝟐 + 𝒃𝒙 + 𝒄  The trinomial must be organized in descending order.  The first term must be positive and different from one (a≠1).  The degree (exponent) of the first term must be twice the degree (exponent) of the second term. EXAMPLE • 2𝑥2 + 11𝑥 + 5 • 7𝑚2 − 23𝑚 + 6
  • 14. 6.TRINOMIAL OF THE FORM 𝐚𝒙𝟐 + 𝒃𝒙 + 𝒄 HOW TO FACTOR?  multiply and divide the trinomial by the main coefficient, that is, a.  In the numerator we carry out the distributive property, in the second term the product is not realized but is left expressed  - The first term is expressed as the square of what was left in parentheses in the second term.  - We apply case 5 (Trinomial of the form x2n+bxn+c) in the numerator.  - We apply case 1 (Common Factor) in the parentheses formed and we simplify the fraction (to eliminate the denominator). EXERCISE 6𝑥2 + 5𝑥 − 4
  • 15. 7. SUM AND DIFFERENCE OF PERFECT CUBES It applies only in binomials, where the first term is positive (the second term can be positive or negative). - It is recognized because the coefficients of the terms are perfect cube numbers
  • 16. 7. SUM AND DIFFERENCE OF PERFECT CUBES HOW TO FACTOR? • two parentheses are opened • In the first parenthesis, put the cube roots of each of the terms and the sign that accompanies them. • In the second parenthesis we write the first term squared, minus (plus) the product of the first by the second, plus the square of the third term. EXERCISES 8𝑥3 − 27𝑦3 27𝑥3 + 125𝑦9