Question

Factor completely.
8u4–8u3–12u2+12u

Answer 1

To factor the expression \(8u^4 - 8u^3 - 12u^2 + 12u\), we will start by factoring out the greatest common factor (GCF) from all the terms.

1. **Identify the GCF**:
The GCF of the coefficients \(8, -8, -12, 12\) is \(4\), and since the smallest power of \(u\) present in all terms is \(u\), the GCF of the entire expression is \(4u\).

2. **Factor out the GCF**:
\[
8u^4 - 8u^3 - 12u^2 + 12u = 4u(2u^3 - 2u^2 - 3u + 3)
\]

3. **Focus on the remaining polynomial**:
Now we need to factor \(2u^3 - 2u^2 - 3u + 3\).

4. **Grouping**:
We can factor by grouping. First, we can group the first two terms and the last two terms:
\[
2u^3 - 2u^2 - 3u + 3 = (2u^3 - 2u^2) + (-3u + 3)
\]
Factor out what is common in each group:
\[
= 2u^2(u - 1) - 3(u - 1)
\]

5. **Factor by grouping**:
Now, we notice that \((u - 1)\) is common in both terms:
\[
= (u - 1)(2u^2 - 3)
\]

6. **Combine everything**:
Now we can write the complete factored form including the GCF we originally factored out:
\[
8u^4 - 8u^3 - 12u^2 + 12u = 4u(u - 1)(2u^2 - 3)
\]

Thus, the completely factored form of the expression \(8u^4 - 8u^3 - 12u^2 + 12u\) is:
\[
\boxed{4u(u - 1)(2u^2 - 3)}
\]