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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].
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