Let's solve each of the exponent multiplication problems step by step, following the rules of exponents:
Rules Used:
1.
Product of Powers Rule: $ a^m \cdot a^n = a^{m+n} $
2.
Negative Exponents: $ a^{-n} = \frac{1}{a^n} $, so we can move terms with negative exponents to the denominator (or vice versa) to make them positive.
3.
Combine like bases by adding their exponents.
4.
Coefficients (numbers) are multiplied separately.
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We'll go through each problem:
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1) $ 3a^3b^5 \cdot 2a^3b^3 $
- Multiply coefficients: $ 3 \cdot 2 = 6 $
- Combine $ a $ terms: $ a^3 \cdot a^3 = a^{3+3} = a^6 $
- Combine $ b $ terms: $ b^5 \cdot b^3 = b^{5+3} = b^8 $
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Answer: $ 6a^6b^8 $
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2) $ 3c^5k^3 \cdot 5c^7k^4 $
- Coefficients: $ 3 \cdot 5 = 15 $
- $ c^5 \cdot c^7 = c^{5+7} = c^{12} $
- $ k^3 \cdot k^4 = k^{3+4} = k^7 $
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Answer: $ 15c^{12}k^7 $
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3) $ -8z^5 \cdot 4z^2y^3 $
- Coefficients: $ -8 \cdot 4 = -32 $
- $ z^5 \cdot z^2 = z^{5+2} = z^7 $
- $ y^3 $ remains as is (only one term)
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Answer: $ -32z^7y^3 $
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4) $ a^2 \cdot a^{-4} $
- Add exponents: $ a^{2 + (-4)} = a^{-2} $
- But we need
positive exponents, so $ a^{-2} = \frac{1}{a^2} $
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Answer: $ \frac{1}{a^2} $
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5) $ b^4c^5 \cdot 4b^2c^8 \cdot 3b^3c^4 $
- Multiply coefficients: $ 1 \cdot 4 \cdot 3 = 12 $
- $ b^4 \cdot b^2 \cdot b^3 = b^{4+2+3} = b^9 $
- $ c^5 \cdot c^8 \cdot c^4 = c^{5+8+4} = c^{17} $
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Answer: $ 12b^9c^{17} $
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6) $ 5a^4 \cdot 2a^2 \cdot 3a $
- Coefficients: $ 5 \cdot 2 \cdot 3 = 30 $
- $ a^4 \cdot a^2 \cdot a^1 = a^{4+2+1} = a^7 $
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Answer: $ 30a^7 $
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7) $ 4x^4y^3 \cdot 2x^5y^7 \cdot 5x^3y^2 $
- Coefficients: $ 4 \cdot 2 \cdot 5 = 40 $
- $ x^4 \cdot x^5 \cdot x^3 = x^{4+5+3} = x^{12} $
- $ y^3 \cdot y^7 \cdot y^2 = y^{3+7+2} = y^{12} $
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Answer: $ 40x^{12}y^{12} $
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8) $ 4z^{-5}w^6 \cdot 4z^{-3}w^2 $
- Coefficients: $ 4 \cdot 4 = 16 $
- $ z^{-5} \cdot z^{-3} = z^{-8} $
- $ w^6 \cdot w^2 = w^{8} $
- So: $ 16z^{-8}w^8 $
- Make exponent positive: $ z^{-8} = \frac{1}{z^8} $
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Answer: $ \frac{16w^8}{z^8} $
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9) $ 3a^2 \cdot 2a^3b^{-4} \cdot 4a^{-7}b^2 $
- Coefficients: $ 3 \cdot 2 \cdot 4 = 24 $
- $ a^2 \cdot a^3 \cdot a^{-7} = a^{2+3-7} = a^{-2} $
- $ b^{-4} \cdot b^2 = b^{-4+2} = b^{-2} $
- So: $ 24a^{-2}b^{-2} $
- Convert to positive exponents:
- $ a^{-2} = \frac{1}{a^2},\quad b^{-2} = \frac{1}{b^2} $
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Answer: $ \frac{24}{a^2b^2} $
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10) $ 3x^5y^3 \cdot 2x^3y^4 \cdot 4x^{-7} $
- Coefficients: $ 3 \cdot 2 \cdot 4 = 24 $
- $ x^5 \cdot x^3 \cdot x^{-7} = x^{5+3-7} = x^1 = x $
- $ y^3 \cdot y^4 = y^{7} $
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Answer: $ 24xy^7 $
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✔ Final Answers (in positive exponents):
1) $ 6a^6b^8 $
2) $ 15c^{12}k^7 $
3) $ -32z^7y^3 $
4) $ \frac{1}{a^2} $
5) $ 12b^9c^{17} $
6) $ 30a^7 $
7) $ 40x^{12}y^{12} $
8) $ \frac{16w^8}{z^8} $
9) $ \frac{24}{a^2b^2} $
10) $ 24xy^7 $
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Parent Tip: Review the logic above to help your child master the concept of multiply and divide exponents worksheet.
