By Cynthia Y. Young
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This Elibron Classics ebook is a facsimile reprint of a 1907 version via Macmillan and Co. , restricted, London. 4th variation
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23 # 1 (-6)-2 Solution: 1 = 24 a. 2-4 = 1 = 3-3 c. 4-3 # 1 1 33 = 1 , a 3b = 1 # = 33 = 27 1 1 3 a 3b 3 1 1 16 = 3 # 24 = = 64 2-4 4 d. - 23 # 1 4 1 = - 23 # (- 6)2 = (- 8)(36) = (-6)-2 Ϫ8 36 ■ YO U R T U R N u b. qxd Evaluate the expressions. a. - 1 5-2 b. 1 # -2 6 3-2 - 288 ■ Answer: a. Ϫ25 b. qxd 8/7/12 20 5:34 PM Page 20 C H A P T E R 0 Prerequisites and Review Now we can evaluate expressions involving positive and negative exponents. How do we evaluate an expression with a zero exponent? We deﬁne any nonzero real number raised to the zero power as 1.
In algebra, however, numbers are often represented by letters (such as x and y), which are called variables. A constant is a ﬁxed (known) number such as 5. A coefﬁcient is the constant that is multiplied by a variable. Quantities within the algebraic expression that are separated by addition or subtraction are referred to as terms. Algebraic Expression DEFINITION An algebraic expression is the combination of variables and constants using basic operations such as addition, subtraction, multiplication, and division.
Solution: (3x)(2x) ϭ 6x2 Multiply the ﬁrst terms. Multiply the outer terms. (3x)(Ϫ5) ϭ Ϫ15x Multiply the inner terms. (1)(2x) ϭ 2x (1)(Ϫ5) ϭ Ϫ5 Multiply the last terms. Add the ﬁrst, outer, inner, and last terms, and identify the like terms. (3x + 1)(2x - 5) = 6x2 - 15x + 2x - 5 ϭ 6x2 Ϫ 13x Ϫ 5 Combine like terms. ■ YO U R T U R N Multiply (2x Ϫ 3)(5x Ϫ 2). qxd 32 8/3/12 11:17 AM Page 32 C H A P T E R 0 Prerequisites and Review Some products of binomials occur frequently in algebra and are given special names.