Chapter Six. Polynomials. Properties of Exponents Algebraic Expressions Addition, Subtraction, and Multiplication Factoring Solving by Factoring


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1 Chapter Six Polynomials Properties of Exponents Algebraic Expressions Addition, Subtraction, and Multiplication Factoring Solving by Factoring
2 Properties of Exponents The properties below form the basis of algebra. Property Rule Example Power of a Product (ab) x = a x b x (2x) 3 = 2 3 x 3 = 8x 3 Power of a Quotient ( a b )x = ax b x (x 2 )3 = x3 2 = x3 3 8 Power of a Power (b x ) y = b xy (x 5 ) 3 = x 15 Product of Powers b x b y = b x+y x 5 x 3 = x 8 Quotient of Powers b x b y = bx y x 5 x 3 = x2 Zero Exponent b 0 = = 1 Negative Exponent b x = 1 b x 23 = = 1 8 These properties are not valid for a negative value to a fractional exponent or zero to the power of zero.
3 Terms and degrees of polynomials Polynomials with up to three terms have special names. Type Number of terms Example Monomial 1 2x 2 Binomial 2 2x 2 + 9x Trinomial 3 2x 2 + 9x + 3 The degree of a polynomial in one variable is the highest exponent. A polynomial of degree n is called an n th degree polynomial, although polynomials of low degree are usually referred to by the names below. Type Degree Example Constant 0 9 Linear 1 9x Quadratic 2 9x 2 Cubic 3 9x 3 A polynomial in a single variable is in standard form if its terms are in order from highest degree to lowest. A term s coefficient is the constant multiplier, such as 2 for 2x 3 or ¾ for 3x3 4. The leading cofficient of an expression is the coefficient of the highestdegree term.
4 Parts of Equations Component Definition Comment y = 8x 3 + cos (x + 5) Equation two expressions set An equation has an y = 8x 3 + cos (x + 5) equal to each other equals sign. Expression one or more terms added together An expression does not have an equals sign. y 8x 3 + cos (x + 5) Term the product of a coefficient and any number of variables and function values An operation on a term applies only once for the whole term. y 8x 3 cos (x + 5) Function Argument an operation that results in a single value (or no value) for any given input the value input into a function Most functions are followed by their argument, which must be in parentheses unless it is only one term. An operation on the function does not affect the argument. cosine x + 5 Many functions in advanced math and computer languages take more than one argument.
5 Solving equations and simplifying fractions To solve an equation, one or more operations must be done to each whole expression. Equation Incorrect Correct Reason ½x = 10 2(½x = 10) 2(½x) = 2(10) An equals sign is not part of an expression and cannot be operated on. ½x = 10(3x) + 1 x = 20(6x) + 2 x = 20(3x) (2x) is a single term, so it should only be multiplied by 2 one time. ½x = sin 3x x = sin 6x x = 2 sin 3x 2x is an argument, not a term. The ½x = 5 x + 4 x = 10 2x + 8 x = 10 x + 4 term to be doubled is sin 3x. To multiply a fraction by 2, multiply by 2 1. Multilplying by 2 2 is multiplying by 1. To reduce a fraction, the same expression must be divided out of every term one time. Fraction Incorrect Correct Reason 4x + 6y 8x + 18z 1 + y 2 + 6z 2x + 3y 4x + 9z The same operation must be done to every term, rather than dividing some 4x x + 18z 2x x + 9z 2x x + 9z terms by 2x and some terms by 6. The 6 under the square root sign is an argument, not a separate term, so it should not be divided separately.
6 Multiplying polynomials To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial. Keep in mind that negative times negative is positive. Example Work Result 4x(x 2 5x + 2) 4x(x 2 ) + 4x(5x) + 4x(2) 4x 3 20x 2 + 8x 4x(x 2 5x + 2) 4x(x 2 ) + 4x(5x) + 4x(2) 4x x 2 8x To multiply a polynomial by a polynomial, as in the example (4x + 3)(x 2 5x + 2), below, multiply each term of one polynomial by each term of the other polynomial, and add the products. Simplify by combining like terms, which are terms that are identical except possibly for the coefficients. Step Work Original problem (4x + 3)(x 2 5x + 2) Multiply 4x(x 2 5x + 2) + 3(x 2 5x + 2) Distribute (4x 3 20x 2 + 8x) + (3x 2 15x + 6) Combine like terms 4x 3 + (20x 2 + 3x 2 ) + (8x 15x) + 6 Write as a polynomial 4x 3 17x 2 7x + 6
7 Special products of binomials The conjugate of a binomial a + b is a b. Multiplying a binomial by itself or by its conjugate is most easily done by following the patterns below. Multiplier Pattern Proof Example same (a + b) 2 = a 2 + 2ab + b 2 (a + b) 2 (3x + 10) 2 = a(a + b) + b(a + b) = a 2 + ab + ab + b 2 = (3x) 2 + 2(3x(10)) = 9x x = a 2 + 2ab + b 2 conjugate (a + b)(a b) = a 2 b 2 (a + b)(a b) = a(a b) + b(a b) = a 2 ab + ab b 2 (a + b) 2 does not equal a 2 + b 2. If you leave out the 2ab term in the middle, you may turn into a goat. (3x + 10)(3x 10) = (3x) = 9x 2 100
8 Factoring Method When to try How to do it Example Common first, before anything else monomial Factor out any monomial that divides evenly into each term. 4x 8 12x 5 + 6x 2 = 2x 2 (2x 6 6x 3 + 3) Special Pattern if it is a difference of squares, a sum or difference of cubes, or a sum of squares plus a middle term Identify the pattern and the values of a and b (see the following chart). Grouping if there are four terms Write it as a sum of two binomials, and factor the two binomials separately. Guess and Check last Guess the factors, and multiply them to see if the product is the polynomial to be factored. 25x 2 81 = (5x + 9)(5x 9) 8x 3 + 6x 2 20x 15 = (8x 3 + 6x 2 ) + (20x 15) = 2x 2 (4x + 3) 5(4x + 3) = (2x 2 5)(4x + 3) 45x x 12 = (9x 2)(5x + 6)? = 45x x 10x 12
9 Special factoring patterns Type Polynomial Factors Example Difference a 2 b 2 (a + b)(a b) 16x 2 9 of Squares = (4x + 3)(4x 3) Perfect a 2 + 2ab + b 2 (a + b) 2 16x 2 24x + 9 Square = (4x 3) 2 Difference a 3 b 3 (a b)(a 2 + ab + b 2 ) 8x 3 27 of Cubes = (2x 3)(4x 2 + 6x + 9) Sum of a 3 + b 3 (a + b)(a 2 ab + b 2 ) 8x Cubes = (2x + 3)(4x 2 6x + 9)
10 Help with guessing for factoring x 2 + bx + c can be factored as (x + m)(x + n), where m + n = b and mn = c. Example m + n = b mn = c Factors x 2 + 9x = = 18 (x + 3)(x + 6) x 2 9x = = 18 (x + 3)(x + 6) ax 2 + bx + c can be rewritten as (ax 2 + mx) + (nx + c), where m + n = b and mn = ac. This new expression can be factored by grouping. Example m + n = b mn = ac Grouping Factors 3x x = = 3 10 (3x 2 + 5x) + (6x + 10) (3x + 5)(x + 2) 8x x = = 8 5 (8x 2 + 4x) + (10x + 5) (4x + 5)(2x + 1)
11 Solving by factoring If a product is zero, then one of the factors must be zero. Therefore, an equation in which one side is factored and the other side is zero can be solved by setting each factor equal to zero, as in the three examples below. Step (x 8)(2x + 10) = 0 x 2 + 9x + 20 = 0 x 2 + 9x + 20 = 2 Set one side x 2 + 9x + 18 = 0 equal to zero Factor one side (x + 4)(x + 5) = 0 (x + 3)(x + 6) = 0 Set each factor x 8 = 0 equal to zero 2x + 10 = 0 Solve x = 8 x = 5 x + 4 = 0 x + 5 = 0 x = 4 x = 5 x + 3 = 0 x + 6 = 0 x = 3 x = 6
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