To find the time traveled, we can use the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Rate}} \]
In this case, the distance traveled is given by:
\[ D = 5w^3 - 7w^2 + 2w + 204 \quad \text{(in miles)} \]
and the rate is given by:
\[ R = w + 3 \quad \text{(in mph)} \]
Now we can substitute these expressions into the time formula:
\[ \text{Time} = \frac{D}{R} = \frac{5w^3 - 7w^2 + 2w + 204}{w + 3} \]
To simplify this expression, we need to perform polynomial long division because we're dividing a polynomial (the distance) by a linear expression (the rate).
Step 1: Set up the long division
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Divide the first term of the numerator by the first term of the denominator:
- \(5w^3 \div w = 5w^2\)
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Multiply the entire divisor \(w + 3\) by \(5w^2\):
- \(5w^2 \cdot (w + 3) = 5w^3 + 15w^2\)
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Subtract \(5w^3 + 15w^2\) from the original polynomial: \[ (5w^3 - 7w^2 + 2w + 204) - (5w^3 + 15w^2) = -22w^2 + 2w + 204 \]
Step 2: Repeat the process
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Divide the first term \(-22w^2\) by the first term \(w\):
- \(-22w^2 \div w = -22w\)
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Multiply \(w + 3\) by \(-22w\):
- \(-22w \cdot (w + 3) = -22w^2 - 66w\)
-
Subtract: \[ (-22w^2 + 2w + 204) - (-22w^2 - 66w) = 68w + 204 \]
Step 3: Repeat again
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Divide \(68w\) by \(w\):
- \(68w \div w = 68\)
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Multiply \(w + 3\) by \(68\):
- \(68 \cdot (w + 3) = 68w + 204\)
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Subtract: \[ (68w + 204) - (68w + 204) = 0 \]
Final result
The polynomial division results in:
\[ \text{Time} = 5w^2 - 22w + 68 \]
Thus, the expression for the time traveled in hours is:
\[ \text{Time} = 5w^2 - 22w + 68 \quad \text{(in hours)} \]