Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To determine which factorizations are correct for the expression [tex]\(\frac{1}{125}x^3 - 64\)[/tex], let's break it down step-by-step.
Given expression:
[tex]\[ \frac{1}{125}x^3 - 64 \][/tex]
We can represent this expression more intuitively:
[tex]\[ \frac{1}{125}x^3 - \frac{64 \times 125}{125} \][/tex]
[tex]\[ = \frac{x^3}{125} - \frac{64 \times 125}{125} \][/tex]
[tex]\[ = \frac{x^3 - 64 \times 125}{125} \][/tex]
[tex]\[ = \frac{x^3 - 8000}{125} \][/tex]
Recognize that [tex]\(8000 = 20^3\)[/tex], so this expression can be rewritten as:
[tex]\[ \frac{x^3 - 20^3}{125} \][/tex]
Recognize this as a difference of cubes:
[tex]\[ x^3 - 20^3 = (x - 20)(x^2 + 20x + 400) \][/tex]
Thus:
[tex]\[ \frac{x^3 - 20^3}{125} = \frac{(x - 20)(x^2 + 20x + 400)}{125} \][/tex]
The expression [tex]\(\frac{1}{125}(x^3 - 64)\)[/tex] can be factored as:
[tex]\[ 64 \left(\frac{x}{20} - 1\right) \left(\left(\frac{x}{20}\right)^2 + \frac{x}{20} + 1\right) \][/tex]
This converts to:
[tex]\[ 64 \times \left(\frac{x}{20} - 1\right) \times \left(\frac{x^2}{400} + \frac{x}{20} + 1\right) \][/tex]
Simplifying each term:
[tex]\[ 64 (0.05x - 1)(0.0025x^2 + 0.05x + 1) \][/tex]
Comparing the given options:
1. [tex]\(\left(\frac{1}{5}x + 4\right)\left(\frac{1}{25}x^2 - 0.8x + 16\right)\)[/tex]
2. [tex]\(\left(\frac{1}{5}x - 4\right)\left(\frac{1}{25}x^2 + 0.8x + 16\right)\)[/tex]
3. [tex]\(\left(\frac{1}{5}x - 4\right)\left(\frac{1}{25}x^2 + 0.8x + 16\right)\)[/tex]
4. [tex]\(\left(\frac{1}{5}x + 4\right)\left(\frac{1}{25}x^2 + 0.8x - 16\right)\)[/tex]
From the correct factorization:
64 (0.05x - 1)(0.0025x^2 + 0.05x + 1)
Notice:
[tex]\[ \left(\frac{1}{5}x - 4\right)(\text{the Polynomial}) = 0.05 * the correct polynomial \][/tex]
So checking the constants in each factorization when comparing to the correct factorization above, let us look at those compared to:
1. [tex]\(\left(\frac{1}{5}x - 4\right)(0.05x^2 + \sum{x}).\)[/tex]
Therefore, there is no match that corresponds to the correct factorization. To summarize:
None of the given factorisations correctly factorizes [tex]\(\frac{1}{125}x^3 - 64\)[/tex].
Given expression:
[tex]\[ \frac{1}{125}x^3 - 64 \][/tex]
We can represent this expression more intuitively:
[tex]\[ \frac{1}{125}x^3 - \frac{64 \times 125}{125} \][/tex]
[tex]\[ = \frac{x^3}{125} - \frac{64 \times 125}{125} \][/tex]
[tex]\[ = \frac{x^3 - 64 \times 125}{125} \][/tex]
[tex]\[ = \frac{x^3 - 8000}{125} \][/tex]
Recognize that [tex]\(8000 = 20^3\)[/tex], so this expression can be rewritten as:
[tex]\[ \frac{x^3 - 20^3}{125} \][/tex]
Recognize this as a difference of cubes:
[tex]\[ x^3 - 20^3 = (x - 20)(x^2 + 20x + 400) \][/tex]
Thus:
[tex]\[ \frac{x^3 - 20^3}{125} = \frac{(x - 20)(x^2 + 20x + 400)}{125} \][/tex]
The expression [tex]\(\frac{1}{125}(x^3 - 64)\)[/tex] can be factored as:
[tex]\[ 64 \left(\frac{x}{20} - 1\right) \left(\left(\frac{x}{20}\right)^2 + \frac{x}{20} + 1\right) \][/tex]
This converts to:
[tex]\[ 64 \times \left(\frac{x}{20} - 1\right) \times \left(\frac{x^2}{400} + \frac{x}{20} + 1\right) \][/tex]
Simplifying each term:
[tex]\[ 64 (0.05x - 1)(0.0025x^2 + 0.05x + 1) \][/tex]
Comparing the given options:
1. [tex]\(\left(\frac{1}{5}x + 4\right)\left(\frac{1}{25}x^2 - 0.8x + 16\right)\)[/tex]
2. [tex]\(\left(\frac{1}{5}x - 4\right)\left(\frac{1}{25}x^2 + 0.8x + 16\right)\)[/tex]
3. [tex]\(\left(\frac{1}{5}x - 4\right)\left(\frac{1}{25}x^2 + 0.8x + 16\right)\)[/tex]
4. [tex]\(\left(\frac{1}{5}x + 4\right)\left(\frac{1}{25}x^2 + 0.8x - 16\right)\)[/tex]
From the correct factorization:
64 (0.05x - 1)(0.0025x^2 + 0.05x + 1)
Notice:
[tex]\[ \left(\frac{1}{5}x - 4\right)(\text{the Polynomial}) = 0.05 * the correct polynomial \][/tex]
So checking the constants in each factorization when comparing to the correct factorization above, let us look at those compared to:
1. [tex]\(\left(\frac{1}{5}x - 4\right)(0.05x^2 + \sum{x}).\)[/tex]
Therefore, there is no match that corresponds to the correct factorization. To summarize:
None of the given factorisations correctly factorizes [tex]\(\frac{1}{125}x^3 - 64\)[/tex].
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.