To factor the expression \( 5v^5 - 5v^4 + 6v^2 - 6v^3 \), we can start by grouping the terms in pairs:
\[ (5v^5 - 5v^4) + (6v^2 - 6v^3) \]
Now, factor out the common factors in each group:
- From the first group \( 5v^5 - 5v^4 \), we can factor out \( 5v^4 \):
\[ 5v^4(v - 1) \]
- From the second group \( 6v^2 - 6v^3 \), we can factor out \( 6v^2 \):
\[ 6v^2(1 - v) \]
Now, rewrite the expression with the factored groups:
\[ 5v^4(v - 1) + 6v^2(1 - v) \]
Notice that \( 1 - v \) can be rewritten as \( -(v - 1) \):
\[ 6v^2(1 - v) = -6v^2(v - 1) \]
Now, we have:
\[ 5v^4(v - 1) - 6v^2(v - 1) \]
Next, factor out the common factor \( (v - 1) \):
\[ (v - 1)(5v^4 - 6v^2) \]
Now, we can factor the quadratic \( 5v^4 - 6v^2 \). First, notice that we can factor out \( v^2 \):
\[ 5v^2(v^2 - \frac{6}{5}) \]
This can also be rewritten in terms of a difference of squares:
\[ 5v^2\left(v^2 - \left(\sqrt{\frac{6}{5}}\right)^2\right) \]
This is in the form of \( a^2 - b^2 \), which factors as \( (a - b)(a + b) \):
Let \( a = v \) and \( b = \sqrt{\frac{6}{5}} \):
\[ v^2 - \left(\sqrt{\frac{6}{5}}\right)^2 = \left(v - \sqrt{\frac{6}{5}}\right)\left(v + \sqrt{\frac{6}{5}}\right) \]
So we can rewrite \( 5v^2(v^2 - \frac{6}{5}) \) as:
\[ 5v^2\left(v - \sqrt{\frac{6}{5}}\right)\left(v + \sqrt{\frac{6}{5}}\right) \]
Combining everything, our completely factored expression becomes:
\[ (v - 1)(5v^2)\left(v - \sqrt{\frac{6}{5}}\right)\left(v + \sqrt{\frac{6}{5}}\right) \]
Thus, the completely factored form of the original expression \( 5v^5 - 5v^4 + 6v^2 - 6v^3 \) is:
\[ 5v^2(v - 1)\left(v - \sqrt{\frac{6}{5}}\right)\left(v + \sqrt{\frac{6}{5}}\right) \]