f the distance traveled is left parenthesis 5 w cubed minus 7 w squared plus 2 w plus 204 right parenthesis miles and the rate is left parenthesis w plus 3 right parenthesis ​mph, write an​ expression, in​ hours, for the time traveled.

1 answer

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

  1. Divide the first term of the numerator by the first term of the denominator:

    • \(5w^3 \div w = 5w^2\)
  2. Multiply the entire divisor \(w + 3\) by \(5w^2\):

    • \(5w^2 \cdot (w + 3) = 5w^3 + 15w^2\)
  3. 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

  1. Divide the first term \(-22w^2\) by the first term \(w\):

    • \(-22w^2 \div w = -22w\)
  2. Multiply \(w + 3\) by \(-22w\):

    • \(-22w \cdot (w + 3) = -22w^2 - 66w\)
  3. Subtract: \[ (-22w^2 + 2w + 204) - (-22w^2 - 66w) = 68w + 204 \]

Step 3: Repeat again

  1. Divide \(68w\) by \(w\):

    • \(68w \div w = 68\)
  2. Multiply \(w + 3\) by \(68\):

    • \(68 \cdot (w + 3) = 68w + 204\)
  3. 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:

\[ \text{Time} = 5w^2 - 22w + 68 \quad \text{(in hours)} \]