- Identify the GCF of the terms \(7a^2b\), \(10a^2b^2\), and \(14a^2b^3\) as \(a^2b\).
- Divide each term of the expression by the GCF: \(\frac{7a^2b}{a^2b} = 7\), \(\frac{10a^2b^2}{a^2b} = 10b\), and \(\frac{14a^2b^3}{a^2b} = 14b^2\).
- Combine the factors to write the equivalent expression: \(a^2b(7 + 10b + 14b^2)\).
- Final Answer: \(\boxed{a^2b(7 + 10b + 14b^2)}\).
Explanation
- Analyze the Expression
The given expression is \(7 a^2 b + 10 a^2 b^2 + 14 a^2 b^3\). To simplify or factor this expression, we first look for the greatest common factor (GCF) among the three terms: \(7 a^2 b\), \(10 a^2 b^2\), and \(14 a^2 b^3\).
- Identify the Greatest Common Factor
We compare the coefficients: 7, 10, and 14. Their greatest common divisor is 1. Next, we examine the variable parts:
- For \(a\): The common factor is \(a^2\), as it appears in every term.
- For \(b\): The lowest power of \(b\) across the terms is \(b^1\), so the common factor is \(b\).
Combining these, the GCF of the expression is \(a^2 b\).
- Factor the Expression
Now, we factor out \(a^2 b\) from each term:
- First term: \(\frac{7 a^2 b}{a^2 b} = 7\)
- Second term: \(\frac{10 a^2 b^2}{a^2 b} = 10 b\)
- Third term: \(\frac{14 a^2 b^3}{a^2 b} = 14 b^2\)
Combining these, the factored form is \(a^2 b(7 + 10 b + 14 b^2)\).
- Conclusion and Final Answer
Comparing our result \(a^2 b(7 + 10 b + 14 b^2)\) with the given options, we see that it matches the second option perfectly. Therefore, the equivalent expression is \(a^2 b(7 + 10 b + 14 b^2)\).
Examples
Factoring expressions is a fundamental skill in algebra with many practical applications. For instance, if you are designing a product and need to calculate costs, you might have an expression for total material cost based on different components. If your formula is \(7a^2b + 10a^2b^2 + 14a^2b^3\), factoring it into \(a^2b(7 + 10b + 14b^2)\) can make it much easier to perform calculations, especially when you need to quickly determine how a change in the variable \(b\) (like the thickness of a material) affects the total cost. This method helps simplify complex models, making them more manageable and easier to interpret.