Let’s solve each problem step by step, using the laws of exponents. We’ll simplify each expression and make sure all exponents are positive.
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Problem 1:
> $ 2x^2 \times 3x^3 \times 2x = $
Step 1: Multiply the coefficients (numbers):
$ 2 \times 3 \times 2 = 12 $
Step 2: Multiply the variables using exponent rule: $ x^a \times x^b = x^{a+b} $
$ x^2 \times x^3 \times x^1 = x^{2+3+1} = x^6 $
✔ Final: $ 12x^6 $
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Problem 2:
> $ 7x^8y^2 \div (x^3y)^2 = $
Step 1: Simplify the denominator first:
$ (x^3y)^2 = x^{3 \cdot 2} y^{1 \cdot 2} = x^6 y^2 $
Step 2: Now divide:
$ \frac{7x^8y^2}{x^6y^2} $
Step 3: Subtract exponents for like bases:
- For x: $ x^{8-6} = x^2 $
- For y: $ y^{2-2} = y^0 = 1 $ → so y disappears
- Coefficient stays 7
✔ Final: $ 7x^2 $
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Problem 3:
> $ 4 \times 2x \times 3x^2y = $
Step 1: Multiply coefficients:
$ 4 \times 2 \times 3 = 24 $
Step 2: Multiply variables:
$ x \times x^2 = x^{1+2} = x^3 $, and we have a y → so $ x^3y $
✔ Final: $ 24x^3y $
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Problem 4:
> $ \frac{4x^2 \times 3x^4}{6x^4} = $
Step 1: Multiply numerator:
$ 4 \times 3 = 12 $, and $ x^2 \times x^4 = x^6 $ → numerator is $ 12x^6 $
Step 2: Divide by denominator:
$ \frac{12x^6}{6x^4} $
Step 3: Simplify coefficients: $ 12 ÷ 6 = 2 $
Simplify exponents: $ x^{6-4} = x^2 $
✔ Final: $ 2x^2 $
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Problem 5:
> $ 4x \times x \times x^2 = $
Step 1: Coefficient is just 4
Step 2: Variables: $ x^1 \times x^1 \times x^2 = x^{1+1+2} = x^4 $
✔ Final: $ 4x^4 $
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Problem 6:
> $ \left( \frac{x^4}{y^2} \right)^3 = $
Apply power to both numerator and denominator:
Numerator: $ (x^4)^3 = x^{4 \cdot 3} = x^{12} $
Denominator: $ (y^2)^3 = y^{2 \cdot 3} = y^6 $
✔ Final: $ \frac{x^{12}}{y^6} $
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Problem 7:
> $ (2x^2)^3 \div 4x^2 = $
Step 1: Expand the cube:
$ (2x^2)^3 = 2^3 \cdot (x^2)^3 = 8x^6 $
Step 2: Divide by $ 4x^2 $:
$ \frac{8x^6}{4x^2} $
Step 3: Coefficients: $ 8 ÷ 4 = 2 $
Exponents: $ x^{6-2} = x^4 $
✔ Final: $ 2x^4 $
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Problem 8:
> $ x^8 \div (x^3)^2 = $
Step 1: Simplify denominator:
$ (x^3)^2 = x^{3 \cdot 2} = x^6 $
Step 2: Divide:
$ \frac{x^8}{x^6} = x^{8-6} = x^2 $
✔ Final: $ x^2 $
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Problem 9:
> $ \frac{x^8 \times (xy^2)^4}{(2x^2)^4} = $
Step 1: Expand numerator:
$ (xy^2)^4 = x^4 y^{8} $ → so numerator becomes:
$ x^8 \cdot x^4 y^8 = x^{12} y^8 $
Step 2: Expand denominator:
$ (2x^2)^4 = 2^4 \cdot (x^2)^4 = 16x^8 $
Step 3: Put together:
$ \frac{x^{12} y^8}{16x^8} $
Step 4: Simplify:
Coefficients: $ \frac{1}{16} $
x: $ x^{12-8} = x^4 $
y: stays $ y^8 $
✔ Final: $ \frac{x^4 y^8}{16} $
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Final Answers:
1. $ 12x^6 $
2. $ 7x^2 $
3. $ 24x^3y $
4. $ 2x^2 $
5. $ 4x^4 $
6. $ \frac{x^{12}}{y^6} $
7. $ 2x^4 $
8. $ x^2 $
9. $ \frac{x^4 y^8}{16} $
Final Answer:
12x^6, 7x^2, 24x^3y, 2x^2, 4x^4, \frac{x^{12}}{y^6}, 2x^4, x^2, \frac{x^4 y^8}{16}
Parent Tip: Review the logic above to help your child master the concept of laws of exponents worksheet 8th grade.