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1 Adding and Subtracting Polynomials Polynomial A monomial or sum of monomials. Binomials and Trinomial are also polynomials. Binomials are sum of two monomials Trinomials are sum of three monomials Degree of MONOMIAL: The sum of all exponents of all variables in the monomial Example: 8x 4 y 2 z Degree = = 7 Degree of POLYNOMIAL: Equals the greatest degree of any term Example: 8bc 2 + 6bc + 5b degree of each monomial Therefore the degree of the polynomial is 3.

2 You Try: Expression Polynomial?/Degree If yes, What kind? binomial 4y 5xz Yes / Yes / 0 monomial 7a 3 + 9b No A negative exponent would put the variable in the denominator 6x 3 + 4x + x 3 Yes / 3 Trinomial (Combine like terms) Standard form of Polynomial degree in descending order For example: 5x 4 + 9x 3 + 7x 2 12x + 2 The degree of the polynomial is 4 and the leading coefficient is 5. When putting polynomials in standard form, be careful to move the sign of the term along with the term: 6 + 5c 2c 3 3c 2 2c 3 3c 2 + 5c is the leading coefficient and the degree is 3

3 Add Polynomials We ve already done this! CLT. Write in Standard Form. Example: (2x 2 + 5x 7) + (3 4x 2 + 6x) Method 1 Method 2 (line up terms) (2x 2 + 5x 7) + (3 4x 2 + 6x) (2x 2 + 5x 7) = 2x x 4 + ( 4x 2 + 6x+3) 2x x 4 Subtract Polynomials Don t forget to distribute the negative sign! CLT. Standard form. Example: (3 2x + 2x 2 ) (4x 5 + 3x 2 ) Method 1 Method 2 Distribute negative sign, then stack (3 2x + 2x 2 ) (4x 5 + 3x 2 ) 3 2x + 2x 2 3 2x + 2x 2 4x + 5 3x x 3x 2 x 2 6x x x 2 Write in standard form: x 2 6x + 8

4 You Try: (5x 2-3x + 4) + (6x - 3x 2-3) (5x 2-3x + 4) + (6x - 3x 2-3) = 2x 2 +3x +1 (y 4 3y + 7) + (2y 3 + 2y - 2y 4-11) (y 4 3y + 7) + (2y 3 + 2y - 2y 4-11) = -y 4 + 2y 3 - y - 4 (4x 3-3x 2 + 6x - 4) (-2x 3 + x 2-2) 4x 3-3x 2 + 6x x 3 - x = 6x 3 4x 2 + 6x - 2 (8y y 2 ) (7 y y) 8y y y 3-12y = y 3 + 5y 2 4y - 17 (7y 2 + 2y 3) + (2 4y + 5y 2 ) (7y 2 + 2y 3) + (2 4y + 5y 2 ) = 12y 2-2y -1 (6y 2 + 8y 4-5y) (9y 4-7y + 2y 2 ) 6y 2 + 8y 4-5y - 9y 4 + 7y - 2y 2 = -y 4 + 4y 2 + 2y (4x 2-2x + 7) + (3x - 7x 2-9) (4x 2-2x + 7) + (3x - 7x 2-9) = -3x 2 + x - 2 (6n n 3 +2n) (4n 3 + 5n 2 ) 6n n 3 +2n 4n + 3-5n 2 = 11n 3 + n 2-2n + 3

5 Multiplying a Polynomial by a Monomial Step 1: Step 2: Use Distributive Property Combine Like Terms as needed Remember: Multiply coefficients Add exponents of terms with same base!! Simplify: 3x 2 (7x 2 x + 4) 21x 4 + 3x 3 12x 2 Remember, multiply coefficients but ADD exponents 2p( 4p 2 + 5p) 5(2p ) 8p p 2 10p CLT! 8p You Try: y(y 12) + y(y + 2) + 25 = 2y(y + 5) 15 y 2 12y + y 2 + 2y + 25 = 2y y 15 2y 2 10y + 25 = 2y y 15 10y + 25 = 10y = 20y 2 = y

6 Multiplying Polynomials Multiplying two binomials: FOIL F + O + I + L Last Inside Outside First F L (y + 8) (y 4) I O F + O + I + L (y)(y) + (y)(-4) + (8)(y) + (8)(-4) TIP: Foil is the same as: You try: 1. Distribute 1 st term 2. Distribute 2 nd term 3. CLT y 2 + ( 4y) + 8y 32 y 2 4y 32 (x + 3)(x 4) (x)(x) (4)(x) + (3)(x) + (3)(-4) = x 2 4x + 3x 12 x 2 x - 12 (4b 5)(3b + 2) 12b 2 + 8b - 15b 10 = 12b 2 7b - 10 (2y 5)(y 6) 2y 2 12y 5y + 30 = 2y 2 17y + 30

7 Multiplying different polynomials: Use distribution: (6x + 5)(2x 2 3x 5) 12x 3 18x 2 30x + 10x 2 15x 25 12x 3 8x 2 45x 25 Simplify the bracket first then distribute: [(t 2 + 3t 8) (t 2 2t + 6)](t 4) (t 2 + 3t 8 t 2 + 2t 6)(t 4) (5t 14)(t 4) 5t 2 20t 14t t 2 34t + 56 Expand equation, then distribute: (3c + 4d + 2) 2 (3c + 4d + 2)(3c + 4d + 2) 9c cd + 6c + 12cd + 16d 2 + 8d + 6c + 8d + 4 9c cd + 12c + 16d + 16d 2 + 4

8 Special Product Polynomials We can always use FOIL or distribution rules to solve these special products, BUT If we recognize that the problem is a special product problem we can use this shortcut: Problem Solution Square of a Sum (a + b) 2 (4x + 7) 2 a= 4x b= 7 a 2 + 2ab + b 2 16x x + 49 Square of a Difference (a b) 2 (5x 4) 2 a 2 2ab + b 2 25x 2 40x + 16 Product of a Sum and Difference a= 5x b= 4 (a + b)(a b) (x + 9)(x 9) a= x b= 9 a 2 b 2 x 2 81 You try: Find the product: (3m + 2n)(3m 2n) = 9m 2 4n 2 (3x + 4y) 2 = 9x xy + 16y 2 (6p 1) 2 = 36p 2 12p + 1 (4c 2)(4c + 2) = 16c 2-4 (8c + 3d) 2 = 64c cd + 9d 2 (a 2b) 2 = a 2 4ab + 4b 2

9 Factor Using GCF s Factor by finding GCF (greatest common factor) Step 1: Find the GCF of the Coefficients Step 2: For variables take the lowest degree per variable (if the variable exists in every term). Step 3: 3: Divide the expression by the GCF Example 1: Factor 27y y Step 1: GCF of 27 and 18 is 9 Step 2: Variable with lowest degree of y 2 and y is y. GCF of both terms is 9y. Step 3: 9y (3y + 2) Example 2: Factor 4a 2 b 8ab 2 + 2ab GCF = 2ab 2ab(-2a - 4b + 1) You need this one. You try: Factor each polynomial: 15 w 3v = 3(5w v) 7u 2 t ut 2 ut = ut(7ut + 21t 1)

10 Factor by Grouping (usually used for polynomials of four or more terms) Step 1: Step 2: Step 3: Group terms with common factors Factor GCF from each group Group Common Factor Example 1: Factor 4qr q + 8r Step 1: (4qr + 8r) + (3q + 6) Step 2: 4r(q + 2) + 3(q + 2) Step 3: Common Factor = (q+2) Final answer: (4r + 3)(q + 2) Example 2: Factor 2mk 12m k (2mk 12m) + (42 7k) 2m(k 6) + 7(6 k) Not quite the same 2m(k 6) + 7( 1)( 6 + k) But we can fix that! 2m (k 6) 7(k 6) Now common (2m 7)(k 6) You try: rn + 5n r 5 rn r + 5n 5 r(n 1) + 5 (n 1) (r + 5)(n - 1) 3np + 15p 4n 20 3p(n + 5) 4(n + 5) (3p 4)(n + 5)

11 Zero Product Property If the product of two factors is 0, then one of the factors MUST be equal to 0. We can use this to solve equations.to SOLVE EQUATIONS, THE EQUATION MUST BE SET EQUAL TO ZERO: Solve and Check: Solve and Check: (2d + 6)(3d 15) = 0 c 2 = 3c CHECK: 2d + 6 = 0 3d 15 = 0 2d = 6 3d = 15 d = 3 d = 5 (2( 3) + 6)(3( 3) 15)? 0 ( 6 + 6)( 9 15)? 0 (0)(24)? 0 0 = 0 (2(5) + 6)(3(5) 15)? 0 (10 + 6)(15 15)? 0 (16)(0)? 0 0 = 0 CHECK: Set equation = to 0 first! c 2 3c = 0 c(c 3) = 0 c = 0 c 3 = 0 c = 3 0 2? 3(0) 0 = 0 3 2? (3)(3) 9 = 9 You try: Solve 8b 2 40b = 0 8b(b 5) = 0 8b = 0 b 5 = 0 b = 0 b = 5

12 Solving ax 2 + bx + c = 0 when a=1 Quadratic Equation ax 2 + bx + c = 0 Standard Form Solving quadratic equations: Must be in standard form - factor & use Zero Product Property Step 1: Identify a, b and c Step 2: Check for GCF and factor out if GCF is NOT 1 (If factoring out a GCF, then identify new a, b and c) Step 3: Find two numbers that multiply to ac and add to b * = (a)(c) + = b Step 4: Divide those two numbers by ax Step 5: Simplify if needed and tilt fraction Example: Factor x 2 + 9x + 20 Step 1: a = 1, b = 9 and c = 20 Step 2: GCF is 1 so move to step 3. Step 3: Find two numbers that multiply to 20 but add to = = 9 Step 4: Divide both numbers by 1x, simplify then tilt. Step 5: 4 1x 5 1x cannot simplify so tilt 1x+4 cannot simplify so tilt 1x+5 Factors are (x + 4)(x + 5)

13 Use Zero Product Property to solve, then check your answer: Example: Solve x 2 + 6x = 27 First set equation equal to 0 then factor and solve. x 2 + 6x 27 = 0 a = 1, b = 6, c = -27 GCF= = = 6 3 tilt x-3 9 1x 1x Factors are (x-3)(x+9) tilt x+9 Use Zero Product Property to solve: x-3 = 0 x+9 = 0 x = 3 x = -9 x = {-9, 3} You try: z 2 3z = 70 z 2 3z 70 = = = (z 10) 7 (z + 7) 1z 1z z 10 = 0 z + 7 = 0 z = 10 z = -7 x 2 + 3x 18 = = (-3) = 3 6 (x + 6) 3 (x - 3) 1x 1x x + 6 = 0 x - 3 = 0 x = -6 x = 3

14 Real World Applications (You must check your answers to make sure they make sense!) A poster has a length that is 6 inches longer than the width. The area of the poster is 616 square inches. Find the width of the poster w w(w + 6) = 616 w+6 w 2 + 6w = 616 w 2 + 6w 616 = 0 a = 1 b = 6 c = ( 22) = ( 22) = w (w+28) 22 1w (w-22) w+28 = 0 w-22 = 0 w=-28 w=22 Doesn t make sense can t have negative width! So Width = 22 The width of the poster is 22 inches.

15 Solving ax 2 + bx + c = 0 when a 1 GREAT NEWS!! We already know how to do this!!! Follow rules for solving when a = 1!! (NOTE: Today we may need to simplify fractions in the process). Remember to look for GCF first! Example: Factor 3x 2 17x + 20 Step 1: a = 3 b = -17 c = 20 Step 2: GCF: 1 Step 3: 5 12 = 3(20) = = 17 Step 4 & 5: 5 3x tilt 3x x = 4 tilt x-4 1x Factors are (3x-5)(x-4) Example: Factor 4n n + 70 a = 4 b = 38 c = 70 GCF = 2 Factor the 2 out: 2(2n n + 35) Factor out the 2 but remember it is part of the solution New a = 2 b = 19 c = 35 GCF = = (2)(35) = = = 7 tilt x+7 5 tilt 2x+5 2x x 2x Factors are 2(x+7)(2x+5) This 2 is what we factored out in step 1

16 Sometimes a Polynomial cannot be factored. In this case = we say that the polynomial is PRIME Factor 4x 2 3x + 5 In this case, we need two numbers that multiply to 20 but add to -3. Factors of None of the factors add to -3 therefore polynomial is prime Solve by using Zero Product Property: Solve 2x 2 + 3x = 5 2x 2 + 3x 5 = 0 a = 2 b = 3 c = -5 GCF: = = 3 2 = 1 tilt x-1 5 tilt 2x+5 2x 1x 2x (x 1)(2x + 5) = 0 x-1 =0 2x+5=0 x=1 x= 5 2

17 You Try: Factor 5x x + 6 a = 5, b = 13, c = 6 2n 2 n 1 a = 2, b = -1, c = = = = = x = 2 x 3 5x 2 = 1 2n n 1 2n (x + 2)(5x + 3) 6x x 8 a = 2, b = 22, c = -8 GCF = 2 (n - 1)(2n + 1) 10y 2 35y + 30 a = 10, b = -35, c = 30 GCF = 5 2(3x x 4) 5(2y 2-7y + 6) a = 3, b = 11, c = -4 a = 2, b = -7, c = = = = = x = 4 x 1 3x 3 2y 4 = 2 2y y 2(x + 4)(3x - 1) 4r 2 r + 7 a = 4, b = -1, c = 7 5(2y - 3)(y - 2) 2x 2 + 3x 5 a = 2, b = 3, c = -5 = 28 + = = = 3 Cannot find two numbers to make the above statements true therefore PRIME 2 = 1 2x x 5 2x (x - 1)(2x + 5)

18 Difference of Squares How to Identify a Difference of Squares (a 2 b 2 ) Must be a binomial Must be a subtraction problem 1 st term must be a perfect square 2 nd term must be a perfect square (page 3 lists perfect squares) Factor a difference of squares (a 2 b 2 ) = (a + b)(a b) Examples: 121x 2 4 a = 11x b = 2 Solution = (11x+2)(11x-2) 81 c 2 a = 9 b = c Solution = (9+c)(9-c) 16x 2 9y 2 a = 4x b = 3y Solution = (4x+3y)(4x-3y) Sometimes we need to make sure we are done factoring: b 4 16 (b 2 + 4)(b 2 4) BUT (b 2 4) is also a difference of squares (b 2 + 4)(b + 2)(b 2) Sometimes we need to use more than 1 technique to factor completely:

19 5x 5 45x 7x x 2 7x 21 Factor GCF: 5x(x 4 9) Difference of Squares: 5x(x 2 + 3)(x 2 3) Factor GCF of 7 7(x 3 + 3x 2 x 3) Grouping 7[(x 3 + 3x 2 ) + ( x 3)] 7[x 2 (x + 3) 1(x + 3)] 7(x 2 1)(x + 3) Difference of 7(x + 1)(x 1)(x + 3) squares To solve That s right Zero Product Property 0 = x = x 2 ( 3 4 )2 0 = (x 3 4 )(x ) 0 = (x 3 4 ) 0 = (x ) x = 3 4 x = 3 4 )

20 Perfect Squares 1 1 = 1 1 = ±1 2 2 = 4 4 = ±2 3 3 = 9 9 = ±3 4 4 = = ±4 5 5 = = ±5 6 6 = = ±6 7 7 = = ±7 8 8 = = ±8 9 9 = = ± = = ± = = ± = = ± = = ± = = ± = = ± = = ± = = ± = = ± = = ± = = ±20

21 Perfect Square Trinomials ax 2 + bx + c 1 st term must be a perfect square 3rd term must be a perfect square 2 nd term must equal (2)( ax 2 )( c) Perfect Square Trinomial? 25x 2 30x + 9 Why? YES 25x 2 is a perfect square 9 is a perfect square 30x=(2)(5x)(3) 49y y + 36 NO 42y 2(7y)(6) 9x x 16 NO - 9x 3 is not a perfect square

22 Factoring Perfect Square Trinomials a 2 + 2ab + b 2 (a + b)(a + b) (a + b) 2 a 2 2ab + b 2 (a b)(a b) (a b) 2 As with all factoring, always check for GCF first! Example: Factor 81x 2 90x + 25 GCF = 1 (9x) 2 2(9x)(5) (5) 2 a b The factors are (9x 5) 2 Example 2: Solve 9x 2 48x = 64 to solve set equal to 0 9x 2 48x + 64 = 0 a = 3x b = 8 (3x 8) 2 = 0 3x 8 = 0 x = 8 3 don t repeat the second factor Square Root Property square root and squared cancel x 2 = 16 (y 6) 2 = 81 x 2 = 16 x = ±4 (y 6) 2 = 81 y 6 = ±9 y 6 = 9 y 6 = -9 y = 15 y = -3

23

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