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Example text

Multiply the coefficients: (–3)(–1) = 3. Each term has a base of y, so your answer has a base of y. Next, add the exponents of y from each term: 9 + 9 = 18. (–3y9)(–y9) = 3y18. 4. Multiply the coefficients: (10)(–2) = –20. Each term has a base of g and a base of h, so your answer has both g and h in it. Next, add the exponents of g from each term: 3 + 5 = 8. The exponent of g in your answer is 8. Finally, add the exponents of h from each term: 5 + 9 = 14. The exponent of h in your answer is 14. (10g3h5)(–2g5h9) = –20g8h14.

Once these definitions are out of the way, we will review how to perform basic operations (addition, subtraction, multiplication, and division) on real numbers, and then show how these same operations can be performed on algebraic quantities. Just as we can write number sentences that add or subtract two numbers, we can write sentences, or expressions, using algebra. These expressions might be given to us in words, so we will learn how to turn these words into expressions that contain numbers and variables.

Whew! Is it really that difficult? Not after you see a few examples. Example (8a5) ÷ (2a3) = First, divide the coefficient of the dividend by the coefficient of the divisor: 8 ÷ 2 = 4. Next, carry the bases of the dividend into your answer. Your answer has a base of a. There are no bases in the dividend that are not in the divisor, and vice versa, so move right on to the next step. For each base that is common to both the dividend and the divisor, subtract the exponent of the base in the divisor from the exponent of the base in the dividend.

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