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# Multiplication and Division of Terms Practice Questions

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Updated on Oct 3, 2011

To review these concepts, go to Multiplication and Division of Terms Study Guide.

## Multiplication and Division of Terms Practice Questions

### Problems

#### Practice 1

1. (5)(6) =
2. (–4)(9) =
3. (24) ÷ (–6) =
4. (–32) ÷ (–8) =

#### Practice 2

1. (–7x10)(4x6) =
2. (14u)(8u4) =
3. (–3y9)(–y9) =
4. (10g3h5)(–2g5h9) =
5. (9a6b11c4)(9a2c2) =
6. (–5j7k7l4)(11l4mn9) =
7. (–10p6)(–12qr10)

#### Practice 3

1. (15v9) ÷ (3v5) =
2. (–63a) ÷ (–7a8) =
3. (30p11q4r2) ÷ (–5q2r3) =
4. (11x3y5) ÷ (11y5z2) =
5. (–42m2n2o5) ÷ (6mn2) =

### Solutions

#### Practice 1

1. Both factors are positive, so your answer is positive: (5)(6) = 30.
2. One factor is positive and one is negative, so your answer is negative: (–4)(9) = –36.
3. One number is positive and one is negative, so your answer is negative: (24) ÷ (–6) = –4.
4. Both numbers are negative, so your answer is positive: (–32) ÷ (–8) = 4.

#### Practice 2

1. Multiply the coefficients: (–7)(4) = –28. Each term has a base of x, so your answer has a base of x. Next, add the exponents ofx from each term: 10 + 6 = 16. (–7x10)(4x6) = –28x16.
2. Multiply the coefficients: (14)(8) = 112. Each term has a base of u, so your answer has a base of u. Next, add the exponents of u from each term.Remember, if a base appears to have no exponent, then it has an exponent of 1. 1 + 4 = 5. (14u)(8u4) = 112u5.
3. 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.
5. Multiply the coefficients: (9)(9) = 81. The first term has bases of a, b, and c (the second term also has bases of a and c), so your answer has bases of a, b, and c. Next, add the exponents of a from each term: 6 + 2 = 8. The exponent of a in your answer is 8. Because the second term does not contain b, the exponent of b in your answer will be the same as the exponent of b in the first term, 11. Finally, add the exponents of c from each term: 4 + 2 = 6. The exponent of c in your answer is 6. (9a6b11c4)(9a2c2) = 81a8b11c6.
6. Multiply the coefficients: (–5)(11) = –55. The first term has bases of j, k, and l and the second term has bases of l, m, and n, so your answer has bases of j, k, l, m, and n. Next, add the exponents of l from each term: 4 + 4 = 8. The exponent of l in your answer is 8. Finally, because the bases j, k, l, m, and n each appear in only one term, their exponents are carried right into your answer: (–5j7k7l4)(11l4mn9) = –55j7k7l8mn9.
7. Multiply the coefficients: (–10)(–12) = 120. The first term has a base of p and the second term has bases of q and r, so your answer has bases of p, q, and r. Because the terms have no bases in common, there are no exponents to add. The exponents of p, q, and r are carried right into your answer. (–10p6)(–12qr10) = 120p6qr10.

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