Lecture 03 special products and factoring

Special Products

1. Grace Christian College Special Products and Factoring

2. Multiply: a Monomial and a Polynomial Multiply each polynomial term by that monomial: Positive numbers – law of distribution 3x 2 (6xy + 3y 2 ) 3x 2 (6xy) + 3x 2 (3y 2 ) 18x 3 y + 9x 2 y 2 Negative coefficients – be careful! -2ab 2 (3bz – 2az + 4z 3 ) -2ab 2 (3bz) – ( -2ab 2 )(2az) + ( -2ab 2 )(4z 3 ) -6ab 2 z + 4a 2 b 2 z – 8ab 2 z 3 Number following polynomial – other distributive law (2x 3 – 3x 2 – 5)( 3x ) 3x (2x 3 ) – ( 3x )(3x 2 ) – ( 3x )(5) 6x 4 – 9x 3 – 15x

3. Multiply: Two Polynomials Horizontal Method: (use the distributive property repeatedly) ( 2a + b )(3a – 2b) = ( 2a + b )(3a) – ( 2a + b )(2b) = 6a 2 + 3ab – (4ab + 2b 2 ) = 6a 2 + 3ab – 4ab – 2b 2 = 6a 2 – ab – 2b 2 Vertical Method: 3x 2 + 2x – 5 4x + 2 Multiply top by rightmost 6x 2 + 4x – 10 Then by term to the left 12x 3 + 8x 2 – 20x____ Lastly, add like terms 12x 3 + 14x 2 – 16x – 10

4. Concept: Similar Binomials Any pair of binomials with matching variable parts When multiplied, they produce a trinomial or binomial Examples:

5. The FOIL Method Useful for Multiplying Two Similar Binomials (x + 8)(x - 5) = x 2 – 5x + 8x – 40 (4t 2 + 5)(3t 2 - 2) = 12t 4 – 8t 2 + 15t 2 – 10 (y – 8)(y 2 + 5) = y 3 + 5y – 8y 2 – 40

6. Practice Using the FOIL Method in Your Head If the polynomials are similar, combine the middle terms (x + 2)(x – 5) = x 2 – 3x – 10 (2y + 3)(4y + 1) = 8y 2 + 14y + 3 (m – 3n)(m – 2n) = m 2 – 5mn + 6n 2 (2x + 3y 2 )(x – 7y 2 ) = 2x 2 – 11xy 2 – 21y 4 (a + b)(c + d) = ac + ad + bc + bd

7. Multiplying 3 or more Polynomials Use same technique as you use for numbers: Multiply any 2 together and simplify the temporary product Multiply that temporary product times any remaining polynomial and simplify -2r(r – 2s)(5r – s) = (use foil on the binomials) -2r(r 2 – 11rs + s 2 ) = (distribute the monomial) -2r 3 – 22r 2 s – 2rs 2

8. The Product of Conjugates ( F + L )( F – L ) = F 2 – L 2 The middle term disappears ONLY when the binomials are conjugates: identical except for different operations Multiplying these is easier than using FOIL! (x + 4)(x – 4) = x 2 – 4 2 = x 2 – 16 (5 + 2w)(5 – 2w) = 25 – 4w 2 (3x 2 – 7)(3x 2 + 7) = 9x 4 – 49 (-4x – 10)(-4x + 10) = 16x 2 – 100 (6 + 4y)(6 – 4x) = use the foil method = 36 – 24x + 24y – 16xy

9. Squaring a Binomial Sum ( F + L )( F + L ) = F 2 + 2 F L + L 2 Square the 1 st term Multiply 1 st times 2 nd , double it, add it Square the 2 nd term Try: (2x + 3) 2 (2x) 2 + 2(6x) + 3 2 4x 2 + 12x + 9 (½x + 5) 2 (½x) 2 + 2(5x/2) + 5 2 ¼x 2 + 5x + 25

10. Squaring a Binomial Difference ( F – L )( F – L ) = F 2 – 2 F L + L 2 Square the 1 st term Multiply 1 st times 2 nd , double it, subtract it Square the 2 nd term and add it Try: (3x - 4) 2 = (3x) 2 – 2(12x) + 4 2 = 9x 2 – 24x + 16 (5a – 2b) 2 = (5a) 2 – 2(10ab) + (2b) 2 = 25a 2 – 20ab + 4b 2

11. Practice Binomial Conjugates and Squares ( F + L )( F – L ) = F 2 – L 2 ( F + L ) 2 = F 2 + 2 FL + L 2 ( F – L ) 2 = F 2 – 2 FL + L 2 (x + 3)(x – 3) = x 2 – 9 (2y – 5)(2y – 5) = 4y 2 – 20y + 25 (m + 3n) 2 = m 2 + 6mn + 9n 2 (2y – 5)(2y + 5) = 4y 2 – 25 (a + b)(a + b) = a 2 + 2ab + b 2 (3x – 7y) 2 = 9x 2 – 42xy + 49y 2

12. Find the product of the following:

13. Find the product 

14. Find the product  Use the FOIL method

15. Find the product 

16. Find the product  .

17. Find the product  Use Dist. Prop. twice

18. Find the product 

19. Factoring by Grouping video

20. Factoring Trinomials Video1 video2

21. Perfect Square Trinomial video

22. Perfect Square Factoring video

23. FACTORING IS THE REVERSE of multiplying.   2 x ² + 9 x − 5 (2 x   ?)( x   ?) (2 x   5)( x   1) or with  x -- (2 x   1)( x   5) ? (2 x − 1)( x + 5) = 2 x ² + 9 x − 5.

24. Problem 1.   Place the correct signs to give the middle term. a)  2 x ² + 7 x − 15 = (2 x − 3)( x + 5) b)  2 x ² − 7 x − 15 = (2 x + 3)( x − 5)  c)  2 x ² − x − 15 = (2 x + 5)( x − 3)  d)  2 x ² − 13 x + 15 = (2 x − 3)( x − 5) 

25. Problem 2.   Factor these trinomials. a)  3 x ² + 8 x + 5  = (3 x + 5)( x + 1) b)  3 x ² + 16 x + 5  = (3 x + 1)( x + 5) c)  2 x ² + 9 x + 7  = (2 x + 7)( x + 1) d)  2 x ² + 15 x + 7  = (2 x + 1)( x + 7) e)  5 x ² + 8 x + 3  = (5 x + 3)( x + 1) f)  5 x ² + 16 x + 3  = (5 x + 1)( x + 3)

26. Problem 3.    Factor these trinomials. a)  2 x ² − 7 x + 5  = (2 x − 5)( x − 1) b)  2 x ² − 11 x + 5  = (2 x − 1)( x − 5) c)  3 x ² + x − 10   = (3 x − 5)( x + 2 ) d)  2 x ² − x − 3   = (2 x − 3)( x + 1) e)  5 x ² − 13 x + 6  = (5 x − 3)( x − 2)

27. Factor completely  6 x 8 + 30 x 7 + 36 x 6 .  To factor completely means to first remove any common factor. 6 x 8 + 30 x 7 + 36 x 6 = 6 x 6 ( x ² + 5 x + 6).  Now continue by factoring the trinomial: = 6 x 6 ( x + 2)( x + 3).

28. Problem 4.   Factor completely.  First remove any common factors. a)   x 3 + 6 x ² + 5 x   = x ( x 2 + 6 x + 5) = x ( x + 5)( x + 1) b)   x 5 + 4 x 4 + 3 x 3   = x 3 ( x 2 + 4 x + 3) = x 3 ( x + 1)( x + 3) c)   x 4 + x 3 − 6 x ²  = x ²( x ² + x − 6) = x ²( x + 3)( x − 2) d)  4 x ² − 4 x − 24  = 4( x ² − x − 6) = 4( x + 2)( x − 3) e)  2 x 3 − 14 x ² − 36 x   = 2 x ( x 2 − 7 x − 18) = 2 x ( x + 2)( x − 9) f)  12 x 10 + 42 x 9 + 18 x 8   = 6 x 8 (2 x ² + 7 x + 3) = 6 x 8 (2 x + 1)( x + 3).

29. Quadratics in different arguments Here is the form of a quadratic trinomial with argument x : ax ² + bx + c . The argument is whatever is being squared.   x is being squared.   x is called the argument.  The argument appears in the middle term. a , b , c are called constants.   In this quadratic, 3 x ² + 2 x − 1, the constants are  3, 2, −1.

30. Now here is a quadratic whose argument is x 3 : 3 x 6 + 2 x 3 − 1. x 6 is the square of x 3 .  

31. Now, since the quadratic with argument x can be factored in this way: 3 x ² + 2 x − 1 = (3 x − 1)( x + 1), then the quadratic with argument x 3 is factored in the same way: 3 x 6 + 2 x 3 − 1 = (3 x 3 − 1)( x 3 + 1). Whenever a quadratic has constants 3, 2, −1, then for any argument, the factoring will be (3 times the argument − 1)(argument + 1).

32. Problem 5.   Multiply out each of the following, which have the same constants, but different argument. c)   ( y 6 + 3)( y 6 − 1)   = y 12 + 2 y 6 − 3 d)   ( x 5 + 3)( x 5 − 1)   = x 10 + 2 x 5 − 3 a)   ( z + 3)( z − 1) = z ² + 2 z − 3 b)   ( y + 3)( y − 1) = y ² + 2 y − 3

33. Problem 6.     Factor each quadratic. a)   x ² − 6 x + 5   = ( x − 1)( x − 5) b)   z ² − 6 z + 5   = ( z − 1)( z − 5) c)   x 8 − 6 x 4 + 5   = ( x 4 − 1)( x 4 − 5) d)   x 10 − 6 x 5 + 5   = ( x 5 − 1)( x 5 − 5) e)   x 6 y 6 − 6 x 3 y 3 + 5   = ( x 3 y 3 − 1)( x 3 y 3 − 5)

34. Problem 7.   Factor each quadratic. a)   x 4 − x ² − 2 = ( x ² − 2)( x ² + 1) b)   y 6 + 2 y 3 − 8 = ( y 3 + 4)( y 3 − 2) c)   z 8 + 4 z 4 + 3 = ( z 4 + 1)( z 4 + 3) d)  2 x 10 + 5 x 5 + 3 = (2 x 5 + 3)( x 5 + 1) e)   x 4 y ² − 3 x ² y − 10 = ( x ² y + 2)( x ² y − 5) f)  cos² x − 5 cos x + 6 = (cos x − 3)(cos x − 2)

35. Additional

36. Reference http://cnx.org/content/m21901/latest/

37. The following are some of the products which occur frequently in Mathematics.