1. Special Products and Factoring 2. Multiply: a Monomial and a PolynomialMultiply each polynomial term by that monomial: Positive numbers law of distribution 3x2(6xy + 3y2) 3x2(6xy) + 3x2(3y2) 18x3y + 9x2y2 Negative coefficients be careful! -2ab2(3bz 2az + 4z3) -2ab2(3bz) (-2ab2)(2az) + (-2ab2)(4z3) -6ab2z + 4a2b2z 8ab2z3 Number following polynomial other distributive law (2x3 3x2 5)(3x) 3x(2x3) (3x)(3x2) (3x)(5) 6x4 9x3 15x 3. Multiply: Two PolynomialsHorizontal Method: (use the distributive property repeatedly)(2a + b)(3a 2b)= (2a + b)(3a) (2a + b)(2b)= 6a2 + 3ab (4ab + 2b2)= 6a2 + 3ab 4ab 2b2= 6a2 ab 2b2Vertical Method: 3x2 + 2x 54x + 2 Multiply top by rightmost6x2 + 4x 10 Then by term to the left12x3 + 8x2 20x____ Lastly, add like terms12x3 + 14x2 16x 10 4. Concept: Similar Binomials Any pair of binomials with matching variable parts When multiplied, they produce a trinomial orbinomial Examples:x and x(3 5) ( 2) x x x x2 (3 5)( 2) 3 10a b and a b ( ) ( 2 )a b a b a ab bx y and x Not Similar( 5 )( 7) 7 5 35( 5 ) ( 7)2 2 ( )( 2 ) 3 22 2 x y x x x xy ya b and b a( ) ( 2 )a b b a a a b bx and x Not Similar(2 3) (2 3)2 3 2x x x x x2 2(2 3)(2 3) 4 3 6 9 2 2 4 2 2( )( 2 ) 2 3 x y and x y ( ) ( )x y x y x y2 2 ( )( ) 5. The FOIL MethodUseful for Multiplying Two Similar Binomials(x + 8)(x - 5) = x2 5x + 8x 40(4t2 + 5)(3t2 - 2) = 12t4 8t2 + 15t2 10(y 8)(y2 + 5) = y3 + 5y 8y2 40 6. PracticeUsing the FOIL Method in Your HeadIf the polynomials are similar, combine the middleterms(x + 2)(x 5) = x2 3x 10(2y + 3)(4y + 1) = 8y2 + 14y + 3(m 3n)(m 2n) = m2 5mn + 6n2(2x + 3y2)(x 7y2) = 2x2 11xy2 21y4(a + b)(c + d) = ac + ad + bc + bd 7. Multiplying 3 or morePolynomials Use same technique as you use for numbers: Multiply any 2 together and simplify the temporaryproduct Multiply that temporary product times anyremaining polynomial and simplify-2r(r 2s)(5r s) = (use foil on the binomials)-2r(r2 11rs + s2) = (distribute the monomial)-2r3 22r2s 2rs2 8. The Product of Conjugates(F + L)(F L) = F2 L2 The middle term disappears ONLY when the binomials areconjugates: identical except for different operations Multiplying these is easier than using FOIL! (x + 4)(x 4) = x2 42 = x2 16 (5 + 2w)(5 2w) = 25 4w2 (3x2 7)(3x2 + 7) = 9x4 49 (-4x 10)(-4x + 10) = 16x2 100 (6 + 4y)(6 4x) = use the foil method= 36 24x + 24y 16xy 9. Squaring a Binomial Sum(F + L)(F + L) = F2 + 2FL + L2Square the 1st termMultiply 1st times 2nd,double it, add itSquare the 2nd termTry:(2x + 3)2(2x)2 + 2(6x) + 324x2 + 12x + 9(x + 5)2(x)2 + 2(5x/2) + 52x2 + 5x + 25 10. Squaring a Binomial Difference(F L)(F L) = F2 2FL + L2Square the 1st termMultiply 1st times 2nd,double it, subtract itSquare the 2nd term and add itTry:(3x - 4)2 =(3x)2 2(12x) + 42 =9x2 24x + 16(5a 2b)2 =(5a)2 2(10ab) + (2b)2 =25a2 20ab + 4b2 11. PracticeBinomial Conjugates and Squares(F + L)(F L) = F2 L2(F + L)2 = F2 + 2FL + L2(F L)2 = F2 2FL + L2 (x + 3)(x 3) = x2 9 (2y 5)(2y 5) = 4y2 20y + 25 (m + 3n)2 = m2 + 6mn + 9n2 (2y 5)(2y + 5) = 4y2 25 (a + b)(a + b) = a2 + 2ab + b2 (3x 7y)2 = 9x2 42xy + 49y2 12. Find the product of the following: 13. Find the product 14. Find the productUse the FOIL method 15. Find the product 16. Find the product . 17. Find the productUse 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.2x + 9x 5(2x ?)(x ?)(2x 5)(x 1)or with x --(2x 1)(x 5) ?(2x 1)(x + 5) = 2x + 9x 5. 24. Problem 1. Place the correct signs to give the middleterm.a) 2x + 7x 15 = (2x 3)(x + 5)b) 2x 7x 15 = (2x + 3)(x 5)c) 2x x 15 = (2x + 5)(x 3)d) 2x 13x + 15 = (2x 3)(x 5) 25. Problem 2. Factor these trinomials.a) 3x + 8x + 5 = (3x + 5)(x + 1)b) 3x + 16x + 5 = (3x + 1)(x + 5)c) 2x + 9x + 7 = (2x + 7)(x + 1)d) 2x + 15x + 7 = (2x + 1)(x + 7)e) 5x + 8x + 3 = (5x + 3)(x + 1)f) 5x + 16x + 3 = (5x + 1)(x + 3) 26. Problem 3. Factor these trinomials.a) 2x 7x + 5 = (2x 5)(x 1)b) 2x 11x + 5 = (2x 1)(x 5)c) 3x + x 10 = (3x 5)(x + 2 )d) 2x x 3 = (2x 3)(x + 1)e) 5x 13x + 6 = (5x 3)(x 2) 27. Factor completely 6x8 + 30x7 + 36x6.To factor completely means to first remove any commonfactor.6x8 + 30x7 + 36x6 = 6x6(x + 5x + 6).Now continue by factoring the trinomial:= 6x6(x + 2)(x + 3). 28. Problem 4. Factor completely. First remove any commonfactors.a) x3 + 6x + 5x = x(x2 + 6x + 5) = x(x + 5)(x + 1)b) x5 + 4x4 + 3x3 = x3(x2 + 4x + 3) = x3(x + 1)(x + 3)c) x4 + x3 6x = x(x + x 6) = x(x + 3)(x 2)d) 4x 4x 24 = 4(x x 6) = 4(x + 2)(x 3)e) 2x3 14x 36x = 2x(x2 7x 18) = 2x(x + 2)(x 9)f) 12x10 + 42x9 + 18x8 = 6x8(2x + 7x + 3) = 6x8(2x + 1)(x + 3). 29. Quadratics in different argumentsHere is the form of a quadratic trinomial with argument x :ax + bx + c.The argument is whatever is being squared. x isbeing squared. x is called the argument. Theargument appears in the middle term.a, b, c are called constants.In this quadratic,3x + 2x 1,the constants are 3, 2, 1. 30. Now here is a quadratic whose argument is x3:3x6 + 2x3 1.x6 is the square of x3. 31. Now, since the quadratic with argument x can be factoredin this way:3x + 2x 1 = (3x 1)(x + 1),then the quadratic with argument x3 is factored in thesame way:3x6 + 2x3 1 = (3x3 1)(x3 + 1).Whenever a quadratic has constants 3, 2, 1, then for anyargument, the factoring will be(3 times the argument 1)(argument + 1). 32. Problem 5. Multiply out each of the following, which havethe same constants, but different argument.a) (z + 3)(z 1) = z + 2z 3b) (y + 3)(y 1) = y + 2y 3c) (y6 + 3)(y6 1) = y12 + 2y6 3d) (x5 + 3)(x5 1) = x10 + 2x5 3 33. Problem 6. Factor each quadratic.a) x 6x + 5 = (x 1)(x 5)b) z 6z + 5 = (z 1)(z 5)c) x8 6x4 + 5 = (x4 1)(x4 5)d) x10 6x5 + 5 = (x5 1)(x5 5)e) x6y6 6x3y3 + 5 = (x3y3 1)(x3y3 5) 34. Problem 7. Factor each quadratic.a) x4 x 2 = (x 2)(x + 1)b) y6 + 2y3 8 = (y3 + 4)(y3 2)c) z8 + 4z4 + 3 = (z4 + 1)(z4 + 3)d) 2x10 + 5x5 + 3 = (2x5 + 3)(x5 + 1)e) x4y 3xy 10 = (xy + 2)(xy 5)f) cosx 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 inMathematics. I a c d ac adII a b a b a b2 22 2 2III a b a b a b a ab b2 2 2IV a b a b a b a ab bV x a x b x 2a b x abVI ax b cx d acx 2ad bc x bdVII a b c d ac bc ad bd... 2. 2...