Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 29: Factoring Special Products

WTAMU > Virtual Math Lab > Intermediate Algebra > Tutorial 29: Factoring Special Products

Answer/Discussion to 1a

First note that there is no GCF to factor out of this polynomial.

Since it is a trinomial, you can try factoring this by trial and error as shown in Tutorial 28: Factoring Trinomials. But if you can recognize that it fits the form of a perfect square trinomial, you can save yourself some time.

*Fits the form of a perfect sq. trinomial
*Factor as the sum of bases squared

There is no more factoring that we can do in this problem.

Note that if we would multiply this out, we would get the original polynomial.

Answer/Discussion to 1b

The first thing that we always check when we are factoring is WHAT?

The GCF. In this case, there is one.

Factoring out the GCF of 2 as was shown in Tutorial 27: The GCF and Factoring by Grouping, we get:

*Factor out the GCF of 2

This fits the form of the difference of cubes. So we will factor using that rule:

*Fits the form of a diff. of two cubes
*Binomial is diff. of bases
*Trinomial is 1st base squared, plus prod. of bases, plus 2nd base squared

There is no more factoring that we can do in this problem.

Note that if we would multiply this out, we would get the original polynomial.

Answer/Discussion to 1c

The first thing that we always check when we are factoring is WHAT?

The GCF. In this case, there is one.

Factoring out the GCF of 5 as was shown in Tutorial 27: The GCF and Factoring by Grouping, we get:

*Factor out the GCF of 5

This fits the form of the difference of squares. So we will factor using that rule:

*Fits the form of a diff. of two squares
*Factor as the prod. of sum and diff. of bases

There is no more factoring that we can do in this problem.

Note that if we would multiply this out, we would get the original polynomial.

Answer/Discussion to 1d

First note that there is no GCF to factor out of this polynomial.

*Group in two's
*Factor out the GCF out of each separate ( )
*Factor out the GCF of (x squared + 3)

There is no more factoring that we can do in this problem.

Note that if we would multiply this out, we would get the original polynomial.

Answer/Discussion to 1e

First note that there is no GCF to factor out of this polynomial.

This is a trinomial that does not fit the form of a perfect square trinomial. Looks like we will have to use trial and error as shown in Tutorial 28: Factoring Trinomials:

*Factor by trial and error

There is no more factoring that we can do in this problem.

Note that if we would multiply this out, we would get the original polynomial.

WTAMU > Virtual Math Lab >Intermediate Algebra >Tutorial 29: Factoring Special Products