Greatest Common Factor and Factoring by Grouping

# Greatest Common Factor and Factoring by Grouping

## Greatest Common Factor and Factoring by Grouping

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##### Presentation Transcript

1. 4∙2 = 8 3∙3∙3∙3 = 81 Greatest Common Factor and Factoring by Grouping Introduction to Factoring 2∙3 = 6 8∙3∙5 = 120

2. Introduction to Factoring Factoring is the opposite of multiplication. Previously you were taught to factor integers. Factoring an integer means to write the integer as a product of two or more integers. In the product 3∙5 = 15 , for example, 3 and 5 are factors of 15 In this chapter you will learn how to factor polynomials. To factor a polynomial means to express the polynomial as a product of two or more polynomials. Factoring 6x² + 5x -4 = (3x +4)(2x – 1) multiplying

3. Greatest Common Factor Let’s factor integers first: 30 1∙30, 3∙10, 2∙15, or 2∙3∙5 The product of 2∙3∙5 consists only of prime numbers and is called the prime factorization. In this first section we will be using the greatest common factor (GCF) . The greatest common factor of two or more integers is the greatest factor common to each integer. It is useful to express the numbers as a product of prime factors

4. Find the GCF • Finding the GCF of a list of Monomials • Step 1: Find the GCF of the numerical coefficients. • Step 2: Find the GCF of the variable factors. • Step 3: The product of the factors found in Steps 1 and 2 is the GCF of the monomials.

5. Identifying the GCF of Two Integers Find the greatest common factor of each pair of integers. Step 1. Find the GCF of the numerical coefficients. Find the prime factorization of each number. 12 and 20 Factors of 12 = 2 ∙ 2 ∙ 3 Factors of 20 = 2 ∙ 2 ∙ 5 3 5 Find the product of common factors The numbers 12 and 20 share two factors of 2. Therefore the greatest common factor is 2 ∙ 2 = 4

6. Find the GCF of 8ab³, 20a²b³, and 28ab² The GCF of the numerical coefficients 8, 20, and 28 is 4,the largest integer that is a factor of each integer. The GCF of the variable factors a, and a² is a, because a is the largest factor common to both powers of a. The variable b², and b³ is b², because b² is the largest factor common to both powers of b.

7. To see this in factored form: 8ab³ = 2·2·2 a b bb 20a²b³ = 2·2·5 a a b b b 28ab² = 2·2·7 a b b GCF = 4ab²

8. Solving a binomial There is a 5 in common in both terms that we could factor out. According to the directions though, we are supposed to factor out a negative real number so let's factor out –5. Yes---it checks! Remember you can always check to see if you've done this step correctly by re-distributing through. Let's check it. Trade the terms places for standard form.

9. Solving a binomial There is a 3y³ in common in both terms that we could factor out. Yes---it checks! Remember you can always check to see if you've done this step correctly by re-distributing through. Let's check it. Trade the terms places for standard form.

10. There is a 2y in common in all three terms that we could factor out. According to the directions though, we are supposed to factor out a negative real number so let's factor out –2. There is also a y common in all three terms. Yes---it checks! Remember you can always check to see if you've done this step correctly by re-distributing through. Let's check it.

11. Steps to Factoring by Grouping • Identify and factor out the GCF from all four terms. • Group the first pair of terms and the second pair of terms. Make sure you always connect the terms by addition. Factor out the GCF from the first pair of terms. Factor out the GCF from the second pair of terms. (Sometimes it is necessary to factor out the opposite of the GCF.) 3. If the two terms share a common binomial factor, factor out the binomial factor

12. Group the first pair of terms and the second pair of terms. Make sure you always connect the terms by addition Identify and factor out the GCF from all four terms. When you see four terms, it is a clue to try factoring by grouping. Look at the first two terms first to see what is in common and then look at the second two terms. These match so let's factor them out There is nothing in common in the second two terms but we want them to match what we have in parentheses in the first two terms so we'll factor out -1

13. Group the first pair of terms and the second pair of terms. Make sure you always connect the terms by addition Identify and factor out the GCF from all four terms. In this case, the GCF is 2w When you see four terms, it is a clue to try factoring by grouping. Look at the first two terms first to see what is in common and then look at the second two terms. These match so let's factor them out af There is a 3 in common in the second two terms but we want them to match what we have in parentheses in the first two terms so we'll factor out -3

14. Factoring Trinomials: Grouping Method Multiply Multiply the binomials. Add the middle terms. Factor Rewrite the middle terms as a sum or difference of terms Factor by grouping

15. Grouping Method Factorax² + bx + c (a ≠ 0) • 1. Multiply the coefficients of the first and last terms(ac). • 2. Find two integers whose product is ac and whose sum is b. (if no pair of integers can be found, then the trinomial cannot be factored further and is a prime polynomial.) • 3. Rewrite the middle terms bx as the sum of two terms whose coefficients are the integers found in Step 2. • 4. Factor by grouping.

16. Factor out the GCF from all the terms. In this case, the GCF is 1 Step 1. The trinomial is written in the form ax² +bx + c. Find the product ac = (2)(6) = 12 TIP: a = 2, b = 7, c = 3 Step 2. List all the factors of ac and search for the pair whose sum equals the value of b. That is, list the factors of 12 and find the pair whose sum equals 7. 1212 1∙12 (-1)(-12) 2∙6 (2)(6) 3∙4 (3)(4) The numbers 3 and 4 satisfy both conditions: 3 ∙ 4 = 12 3 + 4 = 7

17. Step 3. Write the middle term of the trinomial as the sum of two terms whose coefficients are the selected pair of numbers: 3 and 4 Step 4. Factor by grouping.

18. First rewrite the polynomial in the form ax² + bx +c The GCF is 1 Step 1. The trinomial is written in the form ax² +bx + c. Find the product ac = (8)(-3) TIP: a = 8, b = -2, c = -3 Step 2. List all the factors of ac and search for the pair whose sum equals the value of b. That is, list the factors of -24 and find the pair whose sum equals -2. -24-24 -1∙24 (-8)(3) -2∙12 (-12)(2) -4 ∙6 (-6)(4) -3∙8 (-24)(1) The numbers -6 and 4 satisfy both conditions: -6∙ 4 = -24 -6 + 4 = -2

19. Step 3. Write the middle term of the trinomial as the sum of two terms whose coefficients are the selected pair of numbers: -6 and 4 Step 4. Factor by grouping.