Question

Factor Completely
20, n, squared, minus, 60, n, minus, 35
20n
2
−60n−35

Answer 1

To factor the expression \( 20n^2 - 60n - 35 \) completely, follow these steps:

1. **Factor out the greatest common factor (GCF)**:
The coefficients \(20\), \(-60\), and \(-35\) have a GCF of \(5\). So we can factor out \(5\):

\[
20n^2 - 60n - 35 = 5(4n^2 - 12n - 7)
\]

2. **Factor the quadratic \(4n^2 - 12n - 7\)**.
We will use the factoring method known as the AC method or trial and error. First, we need to find two numbers that multiply to \(4(-7) = -28\) and add up to \(-12\).

The factor pairs of \(-28\) are:
- \(1\) and \(-28\)
- \(-1\) and \(28\)
- \(2\) and \(-14\)
- \(-2\) and \(14\)
- \(4\) and \(-7\)
- \(-4\) and \(7\)

The pair \(-14\) and \(2\) adds to \(-12\).

3. **Rewrite** the middle term using the chosen pair:

\[
4n^2 - 14n + 2n - 7
\]

4. **Group the terms**:

\[
(4n^2 - 14n) + (2n - 7)
\]

5. **Factor out the common factors in each group**:

\[
2n(2n - 7) + 1(2n - 7)
\]

6. **Factor out the common binomial**:

\[
(2n - 7)(2n + 1)
\]

7. **Combine everything together**:

Now, we will write the complete factorization including the GCF we factored out initially:

\[
5(2n - 7)(2n + 1)
\]

Thus, the completely factored form of the expression \(20n^2 - 60n - 35\) is:

\[
\boxed{5(2n - 7)(2n + 1)}
\]