Algebraic Expressions A variable A monomial is an expression of the form axk, where a is a real number and k is a nonnegative integer. A binomial

Algebraic Expressions Note that the degree of a polynomial is the highest power of the variable that appears in the polynomial.

Example 1– Adding and Subtracting Polynomials (a) Find the sum (x3 – 6x2 + 2x + 4) + (x3 + 5x2 – 7x). (b) Find the difference (x3 – 6x2 + 2x + 4) – (x3 + 5x2 – 7x). Solution: (a) (x3 – 6x2 + 2x + 4) + (x3 + 5x2 – 7x) = (x3 + x3) + (–6x2 + 5x2) + (2x – 7x) + 4 = 2x3 – x2 – 5x + 4 Group like terms Combine like terms

Example 1 – Solution (b) (x3 – 6x2 + 2x + 4) – (x3 + 5x2 – 7x) cont’d (b) (x3 – 6x2 + 2x + 4) – (x3 + 5x2 – 7x) = x3 – 6x2 + 2x + 4 – x3 – 5x2 + 7x = (x3 – x3) + (– 6x2 – 5x2) + (2x + 7x) + 4 = –11x2 + 9x + 4 Distributive Property Group like terms Combine like terms

Example 3 – Multiplying Polynomials Find the product: (2x + 3) (x2 – 5x + 4) Solution 1: Using the Distributive Property (2x + 3)(x2 – 5x + 4) = 2x(x2 – 5x + 4) + 3(x2 – 5x + 4) = (2x  x2 – 2x  5x + 2x  4) + (3  x2 – 3  5x + 3  4) = (2x3 – 10x2 + 8x) + (3x2 – 15x + 12) Distributive Property Distributive Property Laws of Exponents

Special Product Formulas

Example 4 – Using the Special Product Formulas Use the Special Product Formulas to find each product. (a) (x2 – 2)3 Solution: (a) Substituting A = x2 and B = 2 in Product Formula 5, we get (x2 – 2)3 = (x2)3 – 3(x2)2(2) + 3(x2)(2)2 – 23 = x6 – 6x4 + 12x2 – 8

Example 9 – Recognizing the Form of an Expression Factor each expression. (a) x2 – 2x – 3 (b) (5a + 1)2 – 2(5a + 1) – 3 Solution: (a) x2 – 2x – 3 = (x – 3)(x + 1) (b) This expression is of the form where represents 5a + 1. This is the same form as the expression in part (a), so it will factor as Trial and error

Example 9 – Solution cont’d = (5a – 2)(5a + 2)

Special Factoring Formulas Some special algebraic expressions can be factored using the following formulas. The first three are simply Special Product Formulas written backward.

Example 11 – Factoring Differences and Sums of Cubes Factor each polynomial. (a) 27x3 – 1 (b) x6 + 8 Solution: (a) Using the Difference of Cubes Formula with A = 3x and B = 1, we get 27x3 – 1 = (3x)3 – 13 = (3x – 1)[(3x)2 + (3x)(1) + 12] = (3x – 1) (9x2 + 3x + 1)

Example 11 – Solution cont’d (b) Using the Sum of Cubes Formula with A = x2 and B = 2, we have x6 + 8 = (x2)3 + 23 = (x2 + 2)(x4 – 2x2 + 4)

Example 12 – Recognizing Perfect Squares Factor each trinomial. (a) x2 + 6x + 9 (b) 4x2 – 4xy + y2 Solution: (a) Here A = x and B = 3, so 2AB = 2  x  3 = 6x. Since the middle term is 6x, the trinomial is a perfect square. By the Perfect Square Formula we have x2 + 6x + 9 = (x + 3)2

Example 12 – Solution cont’d (b) Here A = 2x and B = y, so 2AB = 2  2x  y = 4xy. Since the middle term is –4xy, the trinomial is a perfect square. By the Perfect Square Formula we have 4x2 – 4xy + y2 = (2x – y)2

Special Factoring Formulas When we factor an expression, the result can sometimes be factored further. In general, we first factor out common factors, then inspect the result to see whether it can be factored by any of the other methods of this section. We repeat this process until we have factored the expression completely.

Example 13 – Factoring an Expression Completely Factor each expression completely. (a) 2x4 – 8x2 (b) x5y2 – xy6 Solution: (a) We first factor out the power of x with the smallest exponent. 2x4 – 8x2 = 2x2(x2 – 4) = 2x2(x – 2)(x + 2) Common factor is 2x2 Factor x2 – 4 as a difference of squares

Example 13 – Solution cont’d (b) We first factor out the powers of x and y with the smallest exponents. x5y2 – xy6 = xy2(x4 – y4) = xy2(x2 + y2)(x2 – y2 ) = xy2(x2 + y2)(x + y)(x – y) Common factor is xy2 Factor x4 – y4 as a difference of squares Factor x2 – y2 as a difference of squares

Example 15 – Factoring by Grouping Factor each polynomial. (a) x3 + x2 + 4x + 4 (b) x3 – 2x2 – 3x + 6 Solution: (a) x3 + x2 + 4x + 4 = (x3 + x2) + (4x + 4) = x2(x + 1) + 4(x + 1) = (x2 + 4)(x + 1) Group terms Factor out common factors Factor out x + 1 from each term

Example 15 – Solution (b) x3 – 2x2 – 3x + 6 = (x3 – 2x2) – (3x – 6) cont’d (b) x3 – 2x2 – 3x + 6 = (x3 – 2x2) – (3x – 6) = x2(x – 2) – 3(x –2) = (x2 – 3)(x – 2) Group terms Factor out common factors Factor out x – 2 from each term