# Math 101 Study Session Spring 2016 Test 4 Chapter 10, Chapter 11 Chapter 12 Section 1, and Chapter 12 Section 2

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1 Math 101 Study Session Spring 2016 Test 4 Chapter 10, Chapter 11 Chapter 12 Section 1, and Chapter 12 Section 2 April 11, 2016 Chapter 10 Section 1: Addition and Subtraction of Polynomials A monomial is a number, a variable, or a product of numbers and variables. A polynomial is a variable expression in which the terms are monomials. A polynomial of two terms is a binomial. A polynomial of three terms is a trinomial. The degree of a polynomial in one variable is the value of the largest exponent on the variable. The degree of 4x 3 3x 2 + 6x 1 is 3, the degree of 5y 4 2y 3 + y 2 7y + 8 is 4, the degree of a nonzero constant (number) is 0, and the number zero has no degree. Polynomials can be added using a vertical or horizontal format. The opposite of a polynomial is the polynomial with the sign of every term changed, i.e., (x 2 2x + 3) = x 2 + 2x 3. Chapter 10 Section 2: Multiplication of Monomials Rule for Multiplying Exponential Expressions If m and n are integers,then x m x n = x m+n. Rule for Simplifying Powers of Exponential Expressions If m and n are integers, then (x m ) n = x mn Rule for Simplifying Powers of Products If m, n, and p are integers, then (x m y n ) p = x mp y np 1

2 Chapter 10 Section 3: Multiplication of Polynomials FOIL Method (A + B) (C + D) = AC + AD + BC + BD Product of the Sum and Difference of Two Terms The Square of a Binomial (a + b) (a b) = a 2 b 2 (a + b) 2 = (a + b) (a + b) = a 2 + 2ab + b 2 (a b) 2 = (a b) (a b) = a 2 2ab + b 2 Chapter 10 Section 4 Integer Exponents and Scientific Notation Zero as an Exponent If x 0, then x 0 = 1. The expression 0 0 is not defined. Definition of Negative Exponents If n is a positive integer and x 0, then x n = 1 x n and 1 x n. Rules for Dividing Exponential Expressions If m and n are integers and x 0, then xm x n = xm n. Chapter 10 Section 5: Division of Polynomials We divide polynomials using a method similar to the method used for the division of whole numbers. 2

3 Figure 1: Dividing Whole Numbers Figure 2: Dividing Polynomials Chapter 11 Section 1: Common Factors The greatest common factor (GCF) of two or more monomials is the greatest integer and the variable with the smallest exponent that is a factor of each monomial. For example the GCF of x 2, x 4, and x 6 is x 2 because 2 is the smallest exponent of the three monomials. To factor a polynomial means to write the polynomial as a product of other polynomials. Chapter 11 Section 2: Form x 2 + bx + c Factoring Polynomials of the To factor polynomials of the form x 2 + bx + c we ask ourselves What two numbers multiply to give c AND add to give b? In other words, we seek integers m and n such that m n = c and m + n = b. Then the polynomial x 2 + bx + c can be factored as (x + m) (x + n). Thus, x 2 + bx + c = (x + m) (x + n). 3

4 Chapter 11 Section 3: Form ax 2 + bx + c Factoring Polynomials of the One method we can use to factor polynomials of the form ax 2 +bx+c is called the trial and error method. To use the trial and error method, we use the factors of a and the factors of c. Figure 3: Trial and Error Method Another (faster) method we can use to factor polynomials of the form ax 2 + bx + c is called the ac method or the factor by grouping method. To use factor by grouping, we multiply a c and look for two numbers m and n so that a c = m n and m + n = b. Then we will split the middle term into mx + nx so that the polynomial ax 2 + bx + c becomes ax 2 + mx + nx + c. Remember, we can only use factor by grouping on a polynomial that has four terms. Chapter 11 Section 4 Special Factoring Factoring the Difference of Two Squares a 2 b 2 = (a + b)(a b) Factoring a Perfect-Square Trinomial a 2 + 2ab + b 2 = (a + b) (a + b) = (a + b) 2 a 2 2ab + b 2 = (a b) (a b) = (a b) 2 4

5 Factoring the Sum or Difference of Two Perfect Cubes a 3 + b 3 = (a + b) ( a 2 ab + b 2) a 3 b 3 = (a b) ( a 2 + ab + b 2) Certain trinomials that are not quadratic can be expressed as a quadratic trinomials by making suitable variable substitutions. A trinomial is quadratic in form if it can be written as au 2 + bu + c. General Factoring Strategy When factoring a polynomial completely, ask the following questions about the polynomial. 1. Is there a common factor? If so, factor out the GCF. 2. If the polynomial is a binomial, is it the difference of two perfect squares, the sum of two perfect cubes, or the difference of two perfect cubes? If so, factor. 3. If the polynomial is a trinomial, is it a perfect-square trinomial or the product of two binomials? If so, factor. 4. Can the polynomial be factored by grouping? If so, factor. 5. Is each factor non-factorable over the integers? If not, factor. Chapter 11 Section 5: Solving Equations Principle of Zero Products If the product of two (or more) factors is zero, then at least one of the factors must be zero. If a b = 0 then a = 0 or b = 0. Steps in Solving a Quadratic Equation by Factoring 1. Write the equation in standard form. 2. Factor the polynomial. 3. Set each factor equal to zero. 4. Solve each equation for the variable. 5. Check the solutions. Chapter 12 Section 1: Rational Expressions Multiplication and Division of A rational expression is a fraction in which the numerator and the denominator are polynomials. A rational expression is in simplest form when the numerator and the denominator have no common factors other than 1. To simplify the rational expression we use 5

6 the property A C B C = A B. Multiply Rational Expressions To multiply rational expressions we use the property a b c d = ac bd. Divide Rational Expressions To divide rational expressions we use the property a b c d = a b d c Figure 4: Dividing Rational Expressions Using Keep Change Flip Chapter 12 Section 2: Addition and Subtraction of Rational Expressions To find the least common multiple (LCM) of two or more polynomials, first factor each polynomial completely. The LCM is the product of each factor the greatest number of times it occurs in any one factorization. Adding and Subtracting Rational Expressions with Common Denominators To add or subtract rational expressions with the same denominator, we use the properties a b + c b = a + c or a b b c b = a c b Adding and Subtraction Rational Expressions without Common Denominators To add or subtract rational expressions without the same denominator, we find the least common multiple of the denominators and express each fraction in terms of the common denominator. 6

7 Below are some examples for us to try with solutions at the end. 1. Simplify ( 3ab) 2 ( 2ab) Multiply ab (4a 2 3ab 8b 2 ). 3. Multiply (3x 7y) (3x + 5y). 4. Multiply (6x 4y) Simplify 22a4 b 8 c 4 33a 7 b 5 c 6. Write the number in scientific notation. 7. Write the number 7, 320, 000 in scientific notation. 8. Divide 24a2 b + 3ab 21ab 2. 3ab 9. Divide (2x x + 7) (x + 7). 10. Factor 4a a. 11. Factor 3p 3 16p 2 + 5p by grouping. 12. Factor 1 64b Factor 2x 4 13x Factor 7x Solve the equation z 2 + 5z 14 = Multiply y2 + y 20 y 2 + 2y 15 y2 + 5y 24 y 2 + 4y Divide 6a2 y + 3a 2 2x 3 + 4x Simplify 2x 4 4y 2 x + 1 6xy. 12ay + 6a 6x x. 2 7

8 Solutions Solution to 1: ( 3ab) 2 ( 2ab) 3 = ( 3) 2 a 2 b 2 ( 2) 3 a 3 b 3 = 9a 2 b 2 ( 8)a 3 b 3 = 72a 2+3 b 2+3 = 72a 5 b 5 Solution to 2: ab(4a 2 3ab 8b 2 ) = ab(4a 2 ) + ab( 3ab) + ab( 8b 2 ) = 4a 2+1 b 3a 1+1 b 1+1 8ab 1+2 = 4a 3 b 3a 2 b 2 8ab 3 Solution to 3: (3x 7y) (3x + 5y) = (3x)(3x) + (3x)(5y) + ( 7y)(3x) + ( 7y)(5y) = 9x xy 21xy 35y 2 = 9x 2 6xy 35y 2 Solution to 4: (6x 4y) 2 = (6x 4y) (6x 4y) = (6x)(6x) + (6x)( 4y) + ( 4y)(6x) + ( 4y)( 4y) = 36x 2 24xy 24xy + 16y 2 = 36x 2 48xy + 16y 2 Solution to 5: 22a 4 b 8 c 4 33a 7 b 5 c = 2 11b8 5 c a 7 4 = 2b3 c 3 3a 3 Solution to 6: To write in scientific notation, we will move the decimal point to the right until there is only one nonzero number to the left of the decimal point. Hence, we will move the decimal point to the right 6 places. Since we are moving the decimal point to the right, the exponent of 10 will be negative =

9 Solution to 7: To write 7, 320, 000 in scientific notation, first note that 7, 320, 000 = 7, 320, We will move the decimal point to the left until there is only one nonzero number to the left of the decimal point. Hence, we will move the decimal point to the left 6 places. Since we are moving the decimal point to the left, the exponent of 10 will be positive. 7, 320, 000 = Solution to 8: We will use the laws of exponents along with the property a + b c = a c + b c. 24a 2 b + 3ab 21ab 2 3ab = 24a2 b 3ab + 3ab 3ab 21ab2 3ab = 8a b 2 1 = 8a + 1 7b Solution to 9: x + 7 ) 2x x + 7 2x 2 14x Solution to 10: 2x + 1Hence, the solution is 2x + 1. x + 7 x 7 0 4a a 2 + 8a = 4a(a 2 + 3a + 2) = 4a(a + 1)(a + 2) To factor a 2 + 3a + 2, we ask ourselves What two numbers multiply to give 2 and add to give 1? Note that 2 1 = 2 AND = 3. Solution to 11: 3p 3 16p 2 + 5p = p(3p 2 16p + 5) = p(3p 2 15p + p + 5) = p [ (3p 2 15p) + ( p + 5) ] = p [3p(p 5) + 1(p 5)] = p (p 5) (3p 1) Solution to 12: Note: 1 3 = 1 and 64b 3 = (4b) 3. Hence, this polynomial is a difference of cubes. Thus, we can use the difference of cubes formula a 3 b 3 = (a b)(a 2 + ab + b 2 ) with a = 1 and b = 4b. 9

10 1 64b 3 = 1 3 (4b) 3 = (1 4b)( (4b) + (4b) 2 ) = (1 4b)(1 + 4b + 16b 2 ) = (1 4b)(16b 2 + 4b + 1) Solution to 13: This polynomial is quadratic in form. By letting u = x 2, u 2 = (x 2 ) 2 = x 4, the polynomial 2x 4 13x 2 15 becomes 2u 2 13u 15. Once we have factored this polynomial, we will replace u with x 2. 2x 4 13x 2 15 = 2u 2 13u 15 = 2u 2 15u + 2u 15 = (2u 2 15u) + (2u 15) = u(2u 15) + 1(2u 15) = (u + 1)(2u 15) = (x 2 + 1)(2x 2 15) Solution to 14: We will use the difference of squares formula a 2 b 2 = (a b)(a + b). 7x = 7(x 2 81) = 7(x ) = 7(x 9)(x + 9) Solution to 15: To solve this equation, we will factor the left hand side and then use the zero product property. To factor the left hand side, we ask ourselves What two numbers multiply to give -14 and add to give 5? Note that 7 ( 2) = 14 and 7 + ( 2) = 5. z 2 + 5z 14 = 0 (z + 7)(z 2) = 0 z + 7 = 0z 2 = 0 z = 7z = 2 Solution to 16: To multiply two rational expressions, we factor each numerator and denominator and use the property A C B C = A B. 10

11 y 2 + y 20 y 2 + 2y 15 y2 + 5y 24 y 2 + 4y 32 Solution to 17: (y + 5)(y 4) (y + 8)(y 3) = (y + 5)(y 3) (y + 8)(y 4) (y + 5)(y 4)(y + 8)(y 3) = (y + 5)(y 3)(y + 8)(y 4) = 1 6a 2 y + 3a 2 2x 3 + 4x 2 12ay + 6a 6x x = 6a2 y + 3a 2 2 2x 3 + 4x 6x3 + 12x ay + 6a = 3a2 (2y + 1) 2x 2 (x + 2) 6x2 (x + 2) 6a(2y + 1) = 3a2 6x 2 (2y + 1)(x + 2) 6a 2x 2 (x + 2)(2y + 1) = 3a2 6x 2 6a 2x 2 = 3a2 2a = 3a2 1 2 = 3a 2 Solution to 18: First, we must find the LCM of 4y 2 and 6xy. To find the LCM, we use each variable with the largest exponent and the largest product of each prime number. Hence, the LCM of 4y 2 and 6xy is x y 2 = 12xy 2 Note that 4y 2 3x = 12xy 2 and 6xy 2y = 12xy 2. 2x 4 x + 1 4y 2 6xy = 2x 4 4y 2 = 3x 3x x + 1 2y 6xy 2y 2y(x + 1) 12xy 2 3x(2x 4) 12xy 2 = 6x2 12x 12xy 2 2xy + 2y 12xy 2 = 6x2 12x 2xy 2y 12xy 2 = 3x2 6x xy y 6xy 2 11

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