Factor completely. See Example.Example Factor each polynomial completely.a. 8a2b − 4abb. 3...

a.

Step 1:

The terms have a common factor of 4ab, which we factor out.

8a2b − 4ab = 4ab(2a − 1)

Step 2:

There are two terms, but the binomial 2a − 1 is not the difference of two squares or the sum or difference of two cubes.

Step 3:

The factor 2a − 1 cannot be factored further.

b.

Step 1:

Factor out a common factor of 9.

36x2 − 9 = 9(4x2 − 1)

Step 2:

The factor 4x2 − 1 has two terms, and it is the difference of two squares.

9(4x2 − 1) = 9(2x + l)(2x − 1)

Step 3:

No factor can be factored further.

c.

Step 1:

The terms of 2x2 − 5x − 7 contain no common factor other than 1 or − 1.

Step 2:

There are three terms. The trinomial is not a perfect square, so we factor by methods from Section 6.3 or 6.4.

2x2 − 5x − 1 = (2x − 7)(x + 1)

Step 3:

No factor can be factored further.

d.

Step 1:

There is no common factor of all terms of 5p2 + 5 + qp2 + q.

Step 2:

The polynomial has four terms, so try factoring by grouping.

Step 3:

No factor can be factored further.

e.

Step 1:

The terms of 27a3 − b3contain no common factor.

Step 2:

There are two terms and 27a3 − b3 is the difference of cubes.

Step 3:

No factor can be factored further.