Question

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.

Answer 1

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:

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