Definition: 2 (degree) The degree of a term is the sum of the exponents of each variable. Definition: 3 (Polynomial) A polynomial is a sum of terms.

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1 I. Polynomials Anatomy of a polynomial. Definition: (Term) A term is the product of a number and variables. The number is called a coefficient. ax n y m In this example a is called the coefficient, and x and y are the variables with exponents n and m respectively. Definition: (degree) The degree of a term is the sum of the exponents of each variable. Definition: 3 (Polynomial) A polynomial is a sum of terms. Definition: 4 (Degree of Polynomial) The degree of a polynomial is the highest degree of all the terms. Classification of polynomials A polynomial of one term is a monomial. A polynomial of two terms is a binomial. A polynomial of three terms is a trinomial. Adding and Subtracting Polynomials Example: Add or subtract. Write in descending order.. ( 6x + 7x + 3) + (3x x + 4) ( 6x + 7x + 3) + (3x x + 4) 6x + 7x x x + 4 6x + 3x + 7x x x + 6x + 6 Remove parenthesis. Combine like terms. Simplify.. (5y y + ) ( 3y 3 y ) Integer Exponents Properties of Exponents i. a n a m = a n+m a ii. n = a n m a m iii. (a n ) m = a nm iv. (ab) n = a n b n v. a n = and = a n a n a n Simplifying expression. (5y y + ) ( 3y 3 y ) Distribute the negative. 5y y + + 3y 3 + y + 3y 3 + 5y y + y + + 3y 3 + 5y + 4 i. Recall your order of operations. Parenthesis first. ii. Simplify the numerator and denominators separately. Combine like terms. Simplify.

2 iii. Move negative exponents. Example: Simplify using positive exponents.. ( 6p 5 )( 7p 5 ) ( 6p 5 )( 7p 5 ) Reorder. ( 6)( 7)p 5 p 5 Apply property 4p 0. 5r 3 t 6 6r t 8 5r 3 t 6 3. (a 3 )(a 7 b ) 3 6r t 8 Apply property. 5r 6t Multiplying Polynomials. (a 3 )(a 7 b ) 3 Apply property 3. a 3 a b 3 Apply property 5. a a 3 b 3 Apply property. a 8 b 3 Multiplying polynomials by a monomial. Distributive law: Let a, bm and c be real numbers. Then a(b + c) = ab + ac. Example: 3 Multiply. Write in descending order.. 4b (b + 8) 4b (b + 8) 4b 3 + 3b Apply distributive law.. 3y(4y y ) 3y(4y y ) y 3y 3 3y 3 + y Distributive law. Write in descending order. Multiplying Binomials. This is also just a use of the distributive law. However we call it a different name. We use the method called foil. First Outer Inner Last Example: 4 Multiply the binomials (n + 9)(n + 3). First Outer (n + 9) (n + 3) Inner Last

3 n(n) + 3n + 9n + 7 n + n + 7 Example: 5 Multiply. Write in descending order.. (5a + 6)(6a + 5). (x 9y)(8x 3y) Special Products i. (a + b) = a + ab + b ii. (a b)(a + b) = a b Example: 6 Multiply.. ( 3 x 3 5 y)( 3 x y) (5a + 6)(6a + 5) FOIL. 30a + 5a + 36a a + 5a + 30 Simplify. (x 9y)(8x 3y) FOIL. 6x 6xy 7xy + 7y 6x 78xy + 7y ( 3 x 3 5 y)( 3 x y) Simplify.. (3m 4) (a b)(a + b) = a b ( ) ( ) 3 3 x 5 y 9 x 9 5 y Multiplying Polynomials Example: 7 Multiply (x + )(x + 3x ). We will use the distributive law twice. (3m 4) (a + b) = a + ab + b (3m) + (3m)( 4) + (4) 9m 4m + 6 (x + ) (x + 3x ) x 3 + 3x x + x + 6x x 3 + 5x + 5x 3

4 Example: 8 Multiply. Write in descending order.. (x 3)(x 3x + 5) (x 3)(x 3x + 5) Distributive law twice. x 3 6x + 0x 3x + 9x 5 x 3 9x + 9x 5. 3x(x + ) Use (a + b) = a + ab + b, then distribute the 3x Simplify. 3x(x + ) 3x(x + 4x + 4) 3x 3 + x + x Distributive law. 3. [(x + y) 3][(x + y) + 3] Use (a b)(a + b) = a b with a = x + y and b = 3. Dividing Polynomials. Dividing by a monomial. Example: 9 Divide. 9x y + 6xy 3xy. 3xy [(x + y) 3][(x + y) + 3] (x + y) 9 Use (a + b) = a + ab + b. (x) + 4xy + y 9 4x + 4xy + y 9 9x y + 6xy 3xy 3xy 9x y 3xy + 6xy 3xy 3xy 3xy 3x + y Split up. Simplify. Dividing by a polynomial We use long division. Find Multiply Subtract Bring down. Example: 0 Divide.. (8x 3 4x 4x + 5) (x + 3) 4x 8x + 5 x + 3 8x 3 4x 4x + 5 8x 3 x 6x 4x 6x 4x 0x + 5 0x 5 0 4

5 . (x + x + 6) (x + 8) x x+8 x + 8 x + x + 6 x 8x 3x + 6 3x (x 3 x 9) (x 3) There are missing terms so we must use place holders. x + x + 3 x 3 x 3 x + 0x 9 x 3 3x x + 0x x 3x 3x 9 3x (3x x + 7) (x + 6) We have to use a fraction for this one. 3 x x+6 x + 6 3x x + 7 3x 9x 0x + 7 0x

6 II. Scientific Notation Definition: 5 (Scientific Notation) A number is written in scientific notation when it is expressed in the form a 0 n, where a < 0 and n is an integer. Essentially we want one and only one non zero digit to the left of the decimal. Example: Write in scientific notation Ans: Ans: Ans: 5 0 Example: Write in standard notation Ans: Ans: Example: 3 Write in Scientific notation.. (4 0 7 )(3 0 4 ) Multiply the tens together and the numbers together. (4 0 7 )(3 0 4 ) 4(3) Not in scientific notation

7 III. Factoring In general we have a guide to factoring. i. Factor out the GCD. ii. Number of terms. () terms use a special form. a b = (a b)(a + b) a + b = cannot factor. a 3 b 3 = (a b)(a + ab + b ) a 3 + b 3 = (a + b)(a ab + b ) () 3 terms factoring trinomials. Trial and Error. Factor by grouping method. (3) 4 terms is factor by grouping. Example: 4 Factor Completely.. x + 6x + 5 x + 6x + 5 (x + 5)(x + ) Factor the trinomial.. 0a 5ab 5b 0a 5ab 5b GCD. 3. y 3 9y 5(a ab 3b ) Factor the trinomial. 5(a 3b)(a + 3b) y 3 9y y(y 9) y(y 3)(y + 3) GCD. Difference of squares. 4. 9y 4 96z 9y 4 96z Difference of squares. 5. x 3 x x + (3y 4z)(3y 4z) x 3 x x + x (x ) (x ) Factor by grouping. (x )(x ) Can factor x. (x )(x + )(x ) 6. z 3 5p 3 Use special form a 3 b 3 = (a b)(a + ab + b ) With a = z and b = 5p. 7. (x + 3y) 6 Use a b = (a b)(a + b). z 3 5p 3 = (z 5p)(z + 5pz + 5p ) (x + 3y) 6 = (x + 3y 4)(x + 3y + 4) 7

8 IV. Solving Quadratic Equations Definition: 6 (Quadratic Equation) A quadratic equation is an equation of the form ax +bx+c = 0. Theorem: (Zero Product Property) Let a and b be real numbers. If ab = 0 then either a = 0 or b = 0. How can we use this to solve quadratic equations? Steps to Solving i. Set the equation equal to 0. ii. Factor the polynomial. iii. Set each factor equal to zero and solve. Example: 5 Solve.. a = 8a a = 8a a 8a = 8a 8a a 8a = 0 a(a 4) = 0 a = 0 a 4 = 0 a = 0 a = 4 Set equal to zero. Factor.. p(p 4) = 5 p(p 4) = 5 p 4p 5 = 5 5 p 4p 5 = 0 (p 5)(p + ) = 0 p 5 = 0 p + = 0 p = 5 p = Distributive law. Factor. 3. r + 6 = 3r r + 6 = 3r Set equal to 0. r 3r + 6 = 3r 3r r 3r + 6 = 0 Factor. (r )(r 6) = 0 Set each factor equal to 0. r = 0, r 6 = 0 Solve each equation. r =, r = 6 8

9 4. x 3 3x 4x + = 0 x 3 3x 4x + = 0 Factor by grouping. x (x 3) 4(x 3) = 0 (x )(x 3) = 0 Special form on x. (x )(x + )(x 3) = 0 x = 0 x + = 0 x 3 = 0 x = {,, 3} 9

10 V. Rational Expressions. Definition: 7 (Rational Expression) A rational expression is an expression of the form p q where p and q are polynomials with q 0. Steps to reducing a rational expression to lowest terms. i. Factor. ii. Cancel. Write each expression in lowest terms.. x y + y + x z + z xy + xz x y + y + x z + z xy + xz y(x + ) + z(x + ) x(y + z) (y + z)(x + ) x(y + z) x + x Multiplying rational expression. Steps to simplifying. i. Factor. ii. Multiply across. iii. Cancel. Multiply.. 8p p 4 8p 3 Factor. Factor by grouping. Reduce. 6 4 p 4 7 p Cancel.. r + 3 r 4 r 6 6r + 9 r + 3 r 4 r 6 6r + 9 r + 3 (r 4)(r + 4) r 4 3(r + 3) (r + 3)(r 4)(r + 4) (r 4) 3(r + 3) r Factor. Multiply across. Cancel.

11 3. 3z + 5z 9z 9z + 6z + z + 5z + 6 Dividing Rational Expressions Steps to dividing. i. Factor. ii. Multiply by the reciprocal. iii. Cancel. Divide. 8x. x x 5 6x 4 8x x x 5 6x 4 8x x 6x4 5 x 3z + 5z 9z + 6z + 9z z + 5z + 6 (3z )(z + ) (3z + )(3z + ) (3z )(3z + ) (z + 3)(z + ) (3z )(z + )(3z + )(3z + ) (3z )(3z + )(z + 3)(z + ) 3z + z + 3 Factor. Multiply across. Multiply by the reciprocal. cancel. Cancel.. p + 3p + 0 p + p p + 7p + 5 p p 5 p + 3p + 0 p + 7p + 5 p + p p p 5 (p + 5)(p + 4) (p + 5)(p + ) (p + 4)(p 3) (p + 5)(p 3) (p + 5)(p + 4) (p + 4)(p 3) (p + 5)(p 3) (p + 5)(p + ) p + 5 p + Factor. Multiply by the reciprocal. Cancel. Adding/Subtracting Rational expressions. There are two cases. i. Denominators are the same. In this case we just add or subtract the denominators. ii. Denominators are not the same. We must get a common denominator. We use the LCM to get the common denominator. Add or Subtract. x. x 8 6 x 8

12 x x 8 6 x 8 x 6 x 8 (x 8) x 8 The denominators are the same. Reduce.. 4 p 3 p + 3 p 4 p 3 p + 3 p 4 p 3 p + (p 3) 4 p 3 + p + p p + p 3 p 3 p 3 Denominator not the same but we can switch it. The negatives will cancel. Denominators are the same. Reduce. 3. x x 6 3 x + 8x + 6 x x 6 3 Factor the denominators. x + 8x + 6 x (x 4)(x + 4) 3 Need common denominator. (x + 4)(x + 4) LCM = (x 4)(x + 4) x (x + 4) (x 4)(x + 4) (x + 4) 3 (x 4) (x + 4)(x + 4) (x 4) x + 8 (x 4)(x + 4) 3x Combine numerators. (x 4)(x + 4) x + 8 3x + (x 4)(x + 4) x + 0 (x 4)(x + 4)

13 Complex Fractions. Definition: 8 (Complex Fraction) A complex fraction contains a fraction in the numerator and or denominator. To simplify we multiply the numerator and denominator by the LCM of the denominators.. + x x x + 8 x + x x x + 8 LCM = 8x + x x 8x 8x x + 8 (8x) + x(8x) x + (8x) x (x + )x 8x + 8 x 3 + x Multiply top and bottom by LCM. distributive law. Cancel.. 8 x+4 3 x+4 ( 8 ( 3 8 x+4 3 x+4 (x + 4) 8 ) ) x+4 x+4 x+4 x+4 x+4 x+4 x+4 3(x + 4) x+4 x x + x Multiply top and bottom by the LCM. Distributive law. Simplify. 3x 3 3

14 3. x y x y ( ( x y x y x y x y x y x y ) ) x y x y x y x y x y x y x y x y xy x y y x xy(y x) (y x)(y + x) xy (y x) (y x)(y + x) xy y + x Reciprocal for negative exponents. Multiply top and bottom by the LCM. Distributive law. Factor. 4

15 VI. Solving Rational Equations. Steps to solving rational equations. i. Factor the denominators. ii. Multiply both sides by the LCM to clear denominators. iii. Solve. iv. Eliminate any invalid solutions. Solve. 5t. + t = 9 4 5t + t = 9 Clear denominators. ( ) 4 5t 4 + 4(t) = 4(9) Multiply both sides by the LCM. 4 5t + 4t = 36 4 is a valid solution. 9t = 36 t = 4 Solve.. x x x + 3 = 8 x 9 x x x + 3 = 8 x 9 Factor. x x x + 3 = 8 (x 3)(x + 3) LCM = (x 3)(x + 3) Multiply both sides by the LCM. x (x 3)(x + 3) x 3 + (x 3)(x + 3) 4 x + 3 = (x 3)(x + 3) 8 (x 3)(x + 3) x(x + 3) + 4(x 3) = 8 Solve the quadratic equation. x + 3x + 4x = 8 x + 7x = 8 x + 7x 30 = 0 (x + 0)(x 3) = 0 x + 0 = 0 x 3 = 0 x = 0 x = -0 is valid, but 3 is not. x = { 0} 5

16 3. 5 a + 5 = a a a + 5 = a a + 5 LCM = a (a + 5) (a + 5) = a (a + 5) a + 5 a a 0 = a a 5 = a -5 is not valid. No solution. 5 = 3a 5 = a Multiply both sides by LCM. Solve. 6

17 VII. Word Problems.. An airplane maintaining a constant airspeed takes as long to go 450 miles with the wind as it does to go 375 miles against the wind. If the wind is blowing at 5mph, what is the rate of the plane in still air? rate of plane = x rate time distance 450 with wind x x against wind x 5 x Rate of the plane is 65mph. T with = T against 450 x + 5 = 375 x 5 (x + 5)(x 5) = (x + 5)(x 5) x + 5 x 5 450(x 5) = 375(x + 5) 450x 6750 = 375x x 6750 = x = 375 x = 65 7

18 . Working alone Edward can paint a room in 8 hours. Stacey can paint the same room in 6 hours. If they both work together how long will it take. Time to paint together = t work done in hr time total work Edward t t 8 8 Stacey t t t + 6 t = 4 8 t t = 4 3t + 4t = 4 7t = 4 t = 4 7 It takes 4 hours for both to paint together The head gardener can mow the lawn in city park twice as fast as his assistant. Working together they can mow the lawn in hours. How long would it take the head gardener? Head gardener = x Assistant = x work done in hr time total work head x x asst t x x x + x = x + x = x x + x x = x + = x 3 = x It takes the head gardener 3 hours to mow the lawn. 8

19 VIII. Variation Phrase y varies directly as a power of x y varies inversely as a power of x y varies jointly as a power of x and a power of z Formula y = kx n y = k x n y = kx n z m Steps to solving variation problems. i. Determine the equation using the table above. ii. Label the variables according to the problem. iii. Find k using the information. iv. Answer the question. Example: 6 Over a pecified distance, rate varies invereselt with time. If a car on a test track goes a certain distance in half a minute at 60 mph, what is the rate needed to go the same distance in three fourths the rate. Use y = xk with y being the rate and x being the time. Find k using y = 60 and x =. y = kx Plug in y = 60 and x =. 60 = k 30 = k Use k = 30 and x = 3 4 to find y. y = kx Plug in k = 30 and x = 3 4. y = 3 4 (30) y = 40 Example: 7 The weight of a trout varies jointly as its length and the square of its girth. One angler caught a trout that weighed 0.5 lbs and measured 6 in with an 8 in girth. Find the weight of a trout that is in long with a 5 in girth. Use y = kxz with y being weight, x representing length and z representing girth. y = kxz Plug in y = 0.5, x = 6 and z = = k(6)(8) 0.5 = 844k.00 = k y = kxz Plug in k =.00, x = and z = 5. y =.00()(5) y =

20 IX. Function Arithmetic. Definition: 9 Let f(x) and g(x) be functions. Define the following: i. (f + g)(x) = f(x) + g(x) ii. (f g)(x) = f(x) g(x) iii. (fg)(x) = f(x)g(x) ( ) iv. f g (x) = f(x), provided that g(x) = 0. g(x) v. (f g)(x) = f(g(x)) Example: 8 Let f(x) = x 9, g(x) = x, and h(x) = x 3. Find the following.. (f + g)(3). (f h)(x) 3. (gh)(x) (f + g)(3) = f(3) + g(3) = (3) 9 + (3) = = 6 (f h)(x) = f(x) h(x) = x 9 (x 3) = x x = x x 6 (gh)(x) = g(x)h(x) = x(x 3) = x 6x 4. ( f h) () ( ) f () = f() h h() = 9 3 = 5 = 5 5. (f h)(3) (f h)(3) = f(h(3)) h(3) = 3 3 = 0. Replace h(3) with 0. = f(0) = 0 9 = 9 0

21 6. g h(x) g h(x) = g(h(x)) Replace h(x) with x 3. = g(x 3) Plug x 3 into g. = (x 3) = x 6

22 X. Graphing Parabolas. Equation y = x + k y = (x h) y = x Shift Shift up or down k units Shift left or right h units(opposite to the sign). Concave down Example: 9 Graph the parabolas.. y = (x + ) We shift the graph to the left units y = x Shift down units

23 3. y = (x ) + Shift right and up y = (x ) + 4 Shift right, up 4, and concave down

24 XI. Solving A three by three system. Example: 0 Solve the system. x + y z = 5 3x y + z = 5 x + y + 3z = 7 For this example we will use elimination. We will pick a variable to eliminate. The best way to choose is to look at the equations and see which variable will be the easiest to eliminate. For this example it appears that y will be our best candidate. z would also be acceptable. We use the first two equations to eliminate y. To do this we will multiply the second equation by. We get x + y z = 5 + 6x y + z = 0 7x z = 5 We have an equation with two variables. Remember to solve this we need two equations with the same x and z. Therefore we must do this process again. We will use the second and third equations. x + y + 3z = 7 + 6x y + z = 0 4x 4z = We now have two equations with the variables x and z. Now all we have to do is solve the system 7x + z = 5 4x + 4z = We will use elimination to solve this by eliminating x. We will multiply the first equation by 4. 8x 4x = x + 4z = 4x = 48 Solving 4x = 48 we get x =. Now we must solve for the other variables. Since we know what x is, we can plug x = into 4x + 4z = to find z. 4x + 4z = Plug in x =. 4() + 4z = 8 + 4z = 4z = 4 z = Now that we have x = and z =, we will plug these into x + y + 3z = 7 to find y. x + y + 3z = 7 Plug in x = and z =. + y + 3() = 7 y + 5 = 7 y = We have all three variables. Just like the case we write the solution as an ordered triple, or (,, ). 4

25 A system is inconsistent if the system has no solution. A system is called dependent if the system has more than one solution. If we get a false statement the system is inconsistent. If we get a true statement the system is dependent. Example: Solve the following systems... 3x + y + z = 6x + y + z = 9x + 3y + 3z = 3 For this example we will use the first and the second equations to eliminate z. 6x y z = + 6x + y + z = 0 = 0 This is a true statement, and just like before this means the system is dependent. Therefore we get the solution {(x, y, z) : 3x + y + z = } x y + z = x 4y z = 6 x + y z = 5 We will use the first and third equations to eliminate x. x y + z = + x + y z = 5 0 = 7 This is a false statement. Therefore the system is inconsistent and we get no solution or. 5

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