Sure! Let's simplify the expression \((r - 1)(r^2 - 2r + 3)\) using only the coefficients and constants without detailing the letters.
Step 1: Distribute the first term
First, distribute \(r\):
- \(r \cdot 1 = r\)
- \(r \cdot (-2) = -2r\)
- \(r \cdot 3 = 3r\)
Adding these gives:
\[
1r - 2r + 3r
\]
Step 2: Distribute the second term
Now, distribute \(-1\):
- \(-1 \cdot 1 = -1\)
- \(-1 \cdot (-2) = +2\)
- \(-1 \cdot 3 = -3\)
Adding these gives:
\[
-1 + 2 - 3
\]
Step 3: Combine both results
Now combine all the terms together:
- Terms from distributing \(r\): \(r - 2r + 3r\) simplifies to \(2r\)
- Terms from distributing \(-1\): \(-1 + 2 - 3\) simplifies to \(-2\)
Final Result:
So we have:
\[
1r + 0r - 2 \Rightarrow r^3 - 3r^2 + 5r - 3
\]
Thus, the simplified expression is:
\[
\boxed{r^3 - 3r^2 + 5r - 3}
\]
I realize my explanation may still sound complex. Here is a truly numerical form:
Direct Approach with Numerical Representation:
- From \((1)(1) = 1\)
- From \((r)(-2) = -2\)
- From \((r)(3) = 3\)
- From \((-1)(1) = -1\)
- From \((-1)(-2) = 2\)
- From \((-1)(3) = -3\)
Finally, combine:
- Coefficient of \(r^2\): \(-2 - 1 = -3\)
- Coefficient of \(r\): \(3 + 2 = 5\)
The constant term:
- Total = \(-1 + 2 - 3 = -2\)
Thus the final form stays:
\[
r^3 - 3r^2 + 5r - 2
\]
Which was a direct accounting resulted in maintaining the logical terms and please consider rectifying the distribution conditions or mentions. Let me know if you have another numerical focus or format inquiry!