Pregunta
or the binomial completely. \( x^{2}-\frac{1}{25} \) A. \( x^{2}-\frac{1}{25}=\square \) (Factor completely. Simplify your answer. Use integers or fractions for any numbers B. The correct choice below and, if necessary, fill in the prime.
Ask by Mcdonald Pena. in the United States
Jan 25,2025
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The binomial \( x^{2} - \frac{1}{25} \) factors completely as \( \left( x - \frac{1}{5} \right)\left( x + \frac{1}{5} \right) \).
Solución
### Problem A
**Factor the binomial completely:**
Given the expression:
\[ x^{2} - \frac{1}{25} \]
This is a **difference of squares**, which follows the pattern:
\[ a^{2} - b^{2} = (a - b)(a + b) \]
Here, \( a = x \) and \( b = \frac{1}{5} \) because \( \left(\frac{1}{5}\right)^{2} = \frac{1}{25} \).
Applying the difference of squares formula:
\[ x^{2} - \frac{1}{25} = \left( x - \frac{1}{5} \right)\left( x + \frac{1}{5} \right) \]
**Final Factored Form:**
\[ x^{2} - \frac{1}{25} = \left( x - \frac{1}{5} \right)\left( x + \frac{1}{5} \right) \]
### Problem B
**Selection of the Correct Factored Form:**
Based on the factoring above, the correct choice is:
\[ \boxed{\left( x - \dfrac{1}{5} \right)\left( x + \dfrac{1}{5} \right)} \]
If there are multiple-choice options provided, ensure that this factored form matches one of the choices. If the options present the factors differently (e.g., with common denominators or multiplied by constants), adjust accordingly, but generally, the above form is the simplest and fully factored version using integers and fractions.
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To factor the expression \( x^{2}-\frac{1}{25} \), we can recognize that it is a difference of squares. This format can be factored as follows: \( x^{2} - \left(\frac{1}{5}\right)^{2} \). Therefore, it becomes \( (x - \frac{1}{5})(x + \frac{1}{5}) \). Now, let's ensure we have it written correctly. The complete factored form is: \[ x^{2}-\frac{1}{25} = (x - \frac{1}{5})(x + \frac{1}{5}) \]
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