MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §5.3 GCF Grouping

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §5.2 → PolyNomial Multiplication  Any QUESTIONS About HomeWork §5.2 → HW MTH 55

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 3 Bruce Mayer, PE Chabot College Mathematics PolyNomial Factoring Defined  To factor a polynomial is to find an equivalent expression that is a product. An equivalent expression of this type is called a factorization of the polynomial Factoring Breaks an algebraic expression into its simplest pieces –“Simplest”  Smallest Powers

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example  Factoring Monomials  Find three factorizations of 24x 3.  SOLUTION a) 24x 3 = (6  4)(x  x 2 ) = 6x  4x 2 b) 24x 3 = (6  4)(x 2  x) = 6x 2  4x c) 24x 3 = ((−6)(−4))x 3 = (−6)(−4x 3 )

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 5 Bruce Mayer, PE Chabot College Mathematics Greatest Common Factor (GCF)  Find the prime factorization of 105 & 60 Use Factor-Tree    215  35 

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  GCF  Thus  Recognize the Factors that both numbers have in COMMON  The GREATEST Common Factor is the PRODUCT of all the COMMON Factors  In This Case the GCF:

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 7 Bruce Mayer, PE Chabot College Mathematics Examples  GCF  Find the GCF for Monomials: 14p 4 q and 35pq 3  The Prime Factorizations 14p 4 q = 2  7  p  p  p  p  q 35pq 3 = 5  7  p  q  q  q  Thus the GCF = 7  p  q = 7pq

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 8 Bruce Mayer, PE Chabot College Mathematics Examples  GCF  Find the GCF for Three Monomials: 15x 2 30xy 2 57x 3 y  The Prime Factorizations 15x 2 = 3  5  x  x 30xy 2 = 2  3  5  x  y  y 57x 3 y = 3  19  x  x  x  y  Thus the GCF = 3  x = 3x ID the Common Factors

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 9 Bruce Mayer, PE Chabot College Mathematics Factoring When Terms Have a Common Factor  To factor a polynomial with two or more terms of the form ab + ac, we use the distributive law with the sides of the equation switched: ab + ac = a(b + c).  MultiplyFactor  4x(x 2 + 3x − 4) 4x x 2 − 16x  = 4x  x 2 + 4x  3x − 4x  4 = 4x  x 2 + 4x  3x − 4x  4  = 4x x 2 − 16x= 4x(x 2 + 3x − 4)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Factor by Distributive  Factor: 9a − 21  SOLUTION  The prime factorization of 9a is 3  3  a  The prime factorization of 21 is 3  7  The largest common factor is 3.  9a − 21 = 3  3a − 3  7 (UNdist the 3) = 3(3a − 7)  Chk: 3(3a − 7) = 3  3a − 3  7 = 9a − 21 

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Factor by Distributive  Factor: 28x x 3.  SOLUTION  The prime factorization of 28x 6 is  2  2  7  x  x  x  x  x  x  The prime factorization of 32x 3 is  2  2  2  2  2  x  x  x  The largest common factor is 2  2  x  x  x or 4x 3.  28x x 3 = (4x 3  7x ) + (4x 3  8) = 4x 3 (7x 3 + 8)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 12 Bruce Mayer, PE Chabot College Mathematics Factor 12x 5 − 21x x 3  The prime factorization of 12x 5 is 2  2  3  x  x  x  x  x  The prime factorization of 21x 4 is 3  7  x  x  x  x  The prime factorization of 24x 3 is 2  2  2  3  x  x  x  The largest common factor is 3  x  x  x or 3x 3.  12x 5 – 21x x 3 = 3x 3  4x 2 – 3x 3  7x + 3x 3  8 = 3  x  x  x  2  2  x  x = 3  x  x  x  7  x = 3  x  x  x  2  2  2 = 3x 3 (4x 2 – 7x + 8)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Distributive factoring  Factor: 9a 3 b a 2 b 3  SOLUTION  The Prime Factorizations:  The Greatest Common Factor is 9a 2 b 3  Distributing OUT the GCF Produces the factorization: 9a 3 b a 2 b 3 = 9a 2 b 3 (ab + 2)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Distributive factoring  Factor: −4xy + 8xw − 12x  SOLUTION  The Expanded Factorizations −4xy = −4x  y +8xw = − 2  −4x  w − 12x = 3  −4x  Thus the Factored expression: −4xy + 8xw − 12x = −4x(y − 2w + 3)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 15 Bruce Mayer, PE Chabot College Mathematics Factoring Out a Negative GCF  When the coefficient of the term of greatest degree is negative, it is sometimes preferable to factor out the −1 that is understood along with the GCF e.g. Factor Out the GCF for Factor out only the 3. Or factor out the – 3 Both are Correct

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 16 Bruce Mayer, PE Chabot College Mathematics PolyNomial Factoring Tips  Factor out the Greatest Common Factor (GCF), if one exists.  The GCF multiplies a polynomial with the same number of terms as the original polynomial.  Factoring can always be checked by multiplying. Multiplication should yield the original polynomial.

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 17 Bruce Mayer, PE Chabot College Mathematics Factoring by GROUPING  Sometimes algebraic expressions contain a common factor with two or more terms.  Example: Factor x 2 (x + 2) + 3(x + 2)  SOLUTION: The binomial (x + 2) is a factor of BOTH x 2 (x + 2) & 3(x + 2).  Thus, (x + 2) is a common factor; so x 2 (x + 2) + 3(x + 2) = (x + 2)x 2 + (x + 2)3 = (x + 2)(x 2 + 3)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 18 Bruce Mayer, PE Chabot College Mathematics Grouping Game Plan  If a polynomial can be split into groups of terms and the groups share a common factor, then the original polynomial can be factored.  This method, known as factoring by grouping, can be tried on any polynomial with four or more terms

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 19 Bruce Mayer, PE Chabot College Mathematics Examples  Grouping  Factor by grouping. a) 3x 3 + 9x 2 + x + 3 b) 9x 4 + 6x − 27x 3 − 18  Solution a) 3x 3 + 9x 2 + x + 3 = (3x 3 + 9x 2 ) + (x + 3) = 3x 2 (x + 3) + 1(x + 3) = (x + 3)(3x 2 + 1) Don’t Forget the “1”

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 20 Bruce Mayer, PE Chabot College Mathematics Examples  Grouping  Factor by grouping. a) 3x 3 + 9x 2 + x + 3 b) 9x 4 + 6x − 27x 3 − 18  Solution b) 9x 4 + 6x − 27x 3 − 18 = (9x 4 + 6x) + (−27x 3 − 18) = 3x(3x 3 + 2) + (−9)(3x 3 + 2) = (3x 3 + 2)(3x − 9)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Grouping  Factor: y 5 + 5y 3 + 3y  SOLUTION y 5 + 5y 3 + 3y = (y 5 + 5y 3 ) + (3y ) = y 3 (y 2 + 5) + 3(y 2 + 5) = (y 2 + 5) (y 3 + 3) Grouping Factoring each binomial Factoring out the common factor (a BiNomial)

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 22 Bruce Mayer, PE Chabot College Mathematics Factor Factor 4ab + 2ac + 8xb + 4xc  Try grouping terms which have something in common. Often, this can be done in more than one way.  For example or Grp-1Grp-2 a’s & x’s Groupingb’s & c’s Grouping

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 23 Bruce Mayer, PE Chabot College Mathematics Factor Factor 4ab + 2ac + 8xb + 4xc  Next, find the greatest common factor for the polynomial in each set of parentheses.  The GCF for (4ab + 2ac) is 2a  The GCF for (8xb + 4xc) is 4x  The GCF for (4ab + 8xb) is 4b  The GCF for (2ac + 4xc) is 2c Grouping Set-1Grouping Set-2

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 24 Bruce Mayer, PE Chabot College Mathematics Factor Factor 4ab + 2ac + 8xb + 4xc  Write each of the polynomials in parentheses as the product of the GCF and the remaining polynomial  Apply the distributive property to any common factors

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 25 Bruce Mayer, PE Chabot College Mathematics Factor Factor 4ab + 2ac + 8xb + 4xc  Examine the Factorizations  Notice that it did not matter how the terms were originally grouped, the factored forms of the polynomials are IDENTICAL

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 26 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §5.3 Exercise Set 22, 32, 52, 56, 68, 84  Factor by Grouping

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 27 Bruce Mayer, PE Chabot College Mathematics All Done for Today Factoring 4-Term Polynomials

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 28 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 29 Bruce Mayer, PE Chabot College Mathematics

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 30 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 31 Bruce Mayer, PE Chabot College Mathematics

MTH55_Lec-22_sec_5-3_GCF-n-Grouping.ppt 32 Bruce Mayer, PE Chabot College Mathematics Factor Factor 4ab + 2ac + 8xb + 4xc  Divide each polynomial in parentheses by the GCF