Question

Factor completely.
2r5–r4–4r3+2r2

Answer 1

To factor the expression \(2r^5 - r^4 - 4r^3 + 2r^2\), we start by looking for common factors in all the terms.

1. First, notice that each term has a power of \(r\). The smallest power of \(r\) is \(r^2\). We will factor \(r^2\) out from the entire expression:

\[
2r^5 - r^4 - 4r^3 + 2r^2 = r^2(2r^3 - r^2 - 4r + 2)
\]

2. Next, we need to factor the polynomial \(2r^3 - r^2 - 4r + 2\). We can apply factoring by grouping:

- Group the first two terms and the last two terms:

\[
(2r^3 - r^2) + (-4r + 2)
\]

- Now factor out common factors in each group:

\[
r^2(2r - 1) - 2(2r - 1)
\]

3. Notice that \((2r - 1)\) is a common factor:

\[
(2r - 1)(r^2 - 2)
\]

4. Now, substitute back into our expression:

\[
r^2(2r - 1)(r^2 - 2)
\]

5. Finally, check if \(r^2 - 2\) can be factored further. It can be factored as a difference of squares:

\[
r^2 - 2 = (r - \sqrt{2})(r + \sqrt{2})
\]

6. Therefore, the completely factored form of the expression is:

\[
r^2(2r - 1)(r - \sqrt{2})(r + \sqrt{2})
\]

Thus, the final answer is:

\[
\boxed{r^2(2r - 1)(r - \sqrt{2})(r + \sqrt{2})}
\]