Learning Objectives
By the end of this section, you will be able to:
 Factor perfect square trinomials
 Factor differences of squares
 Factor sums and differences of cubes
 Choose method to factor a polynomial completely
Be Prepared 7.11
Before you get started, take this readiness quiz.
Simplify: ${(12x)}^{2}.$
If you missed this problem, review Example 6.23.
Be Prepared 7.12
Multiply: ${(m+4)}^{2}.$
If you missed this problem, review Example 6.47.
Be Prepared 7.13
Multiply: ${(p9)}^{2}.$
If you missed this problem, review Example 6.48.
Be Prepared 7.14
Multiply: $(k+3)(k3).$
If you missed this problem, review Example 6.52.
The strategy for factoring we developed in the last section will guide you as you factor most binomials, trinomials, and polynomials with more than three terms. We have seen that some binomials and trinomials result from special products—squaring binomials and multiplying conjugates. If you learn to recognize these kinds of polynomials, you can use the special products patterns to factor them much more quickly.
Factor Perfect Square Trinomials
Some trinomials are perfect squares. They result from multiplying a binomial times itself. You can square a binomial by using FOIL, but using the Binomial Squares pattern you saw in a previous chapter saves you a step. Let’s review the Binomial Squares pattern by squaring a binomial using FOIL.
The first term is the square of the first term of the binomial and the last term is the square of the last. The middle term is twice the product of the two terms of the binomial.
The trinomial 9x^{2} + 24 +16 is called a perfect square trinomial. It is the square of the binomial 3x+4.
We’ll repeat the Binomial Squares Pattern here to use as a reference in factoring.
Binomial Squares Pattern
If a and b are real numbers,
When you square a binomial, the product is a perfect square trinomial. In this chapter, you are learning to factor—now, you will start with a perfect square trinomial and factor it into its prime factors.
You could factor this trinomial using the methods described in the last section, since it is of the form ax^{2} + bx + c. But if you recognize that the first and last terms are squares and the trinomial fits the perfect square trinomials pattern, you will save yourself a lot of work.
Here is the pattern—the reverse of the binomial squares pattern.
Perfect Square Trinomials Pattern
If a and b are real numbers,
To make use of this pattern, you have to recognize that a given trinomial fits it. Check first to see if the leading coefficient is a perfect square, ${a}^{2}$. Next check that the last term is a perfect square, ${b}^{2}$. Then check the middle term—is it twice the product, 2ab? If everything checks, you can easily write the factors.
Example 7.42
How to Factor Perfect Square Trinomials
Factor: $9{x}^{2}+12x+4$.
Solution
Try It 7.83
Factor: $4{x}^{2}+12x+9$.
Try It 7.84
Factor: $9{y}^{2}+24y+16$.
The sign of the middle term determines which pattern we will use. When the middle term is negative, we use the pattern ${a}^{2}2ab+{b}^{2}$, which factors to ${\left(ab\right)}^{2}$.
The steps are summarized here.
How To
Factor perfect square trinomials.
$\begin{array}{ccccccc}\mathbf{\text{Step 1.}}\phantom{\rule{0.2em}{0ex}}\text{Does the trinomial fit the pattern?}\hfill & & & \hfill {a}^{2}+2ab+{b}^{2}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{a}^{2}2ab+{b}^{2}\hfill \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Is the first term a perfect square?}\hfill & & & \hfill {\left(a\right)}^{2}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{\left(a\right)}^{2}\hfill \\ \phantom{\rule{4em}{0ex}}\text{Write it as a square.}\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Is the last term a perfect square?}\hfill & & & {\left(a\right)}^{2}\phantom{\rule{4.5em}{0ex}}{\left(b\right)}^{2}\hfill & & & \phantom{\rule{2em}{0ex}}{\left(a\right)}^{2}\phantom{\rule{4.5em}{0ex}}{\left(b\right)}^{2}\hfill \\ \phantom{\rule{4em}{0ex}}\text{Write it as a square.}\hfill & & & & & & \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Check the middle term. Is it}\phantom{\rule{0.2em}{0ex}}2ab?\hfill & & & {\left(a\right)}^{2}{}_{\text{\u2198}}\underset{2\xb7a\xb7b}{}{}_{\text{\u2199}}{\left(b\right)}^{2}\hfill & & & \phantom{\rule{2em}{0ex}}{\left(a\right)}^{2}{}_{\text{\u2198}}\underset{2\xb7a\xb7b}{}{}_{\text{\u2199}}{\left(b\right)}^{2}\hfill \\ \mathbf{\text{Step 2.}}\phantom{\rule{0.2em}{0ex}}\text{Write the square of the binomial.}\hfill & & & \hfill {\left(a+b\right)}^{2}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}{\left(ab\right)}^{2}\hfill \\ \mathbf{\text{Step 3.}}\phantom{\rule{0.2em}{0ex}}\text{Check by multiplying.}\hfill & & & & & & \end{array}$
We’ll work one now where the middle term is negative.
Example 7.43
Factor: $81{y}^{2}72y+16$.
Solution
The first and last terms are squares. See if the middle term fits the pattern of a perfect square trinomial. The middle term is negative, so the binomial square would be ${(ab)}^{2}$.
Are the first and last terms perfect squares?  
Check the middle term.  
Does is match ${(ab)}^{2}$? Yes.  
Write the square of a binomial.  
Check by mulitplying.  
${(9y4)}^{2}$  
${\left(9y\right)}^{2}2\cdot 9y\cdot 4+{4}^{2}$  
$81{y}^{2}72y+16\u2713$ 
Try It 7.85
Factor: $64{y}^{2}80y+25$.
Try It 7.86
Factor: $16{z}^{2}72z+81$.
The next example will be a perfect square trinomial with two variables.
Example 7.44
Factor: $36{x}^{2}+84xy+49{y}^{2}$.
Solution
Test each term to verify the pattern.  
Factor.  
Check by mulitplying.  
${(6x+7y)}^{2}$  
${\left(6x\right)}^{2}+2\cdot 6x\cdot 7y+{\left(7y\right)}^{2}$  
$36{x}^{2}+84xy+49{y}^{2}\u2713$ 
Try It 7.87
Factor: $49{x}^{2}+84xy+36{y}^{2}$.
Try It 7.88
Factor: $64{m}^{2}+112mn+49{n}^{2}$.
Example 7.45
Factor: $9{x}^{2}+50x+25$.
Solution
$\begin{array}{}\\ \hfill 9{x}^{2}+50x+25\hfill \end{array}$  
Are the first and last terms perfect squares?  $\begin{array}{}\\ \\ \hfill {\left(3x\right)}^{2}\phantom{\rule{3em}{0ex}}{\left(5\right)}^{2}\hfill \end{array}$ 
Check the middle term—is it $2ab?$  $\begin{array}{}\\ \\ \hfill {\left(3x\right)}^{2}{}_{\text{\u2198}}\underset{\underset{30x}{2\left(3x\right)\left(5\right)}}{\text{}}{}_{\text{\u2199}}{\left(5\right)}^{2}\hfill \end{array}$ 
No! $30x\ne 50x$  This does not fit the pattern! 
Factor using the “ac” method.  $9{x}^{2}+50x+25$ 
$\begin{array}{}\\ \\ \\ \text{Notice:}\phantom{\rule{0.2em}{0ex}}\begin{array}{c}\hfill ac\hfill \\ \hfill 9\xb725\hfill \\ \hfill 225\hfill \end{array}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\begin{array}{}\\ \hfill 5\xb745=225\hfill \\ \hfill 5+45=50\hfill \end{array}\hfill \end{array}$  
Split the middle term. Factor by grouping. 
$\begin{array}{}\\ \hfill 9{x}^{2}+5x+45x+25\hfill \\ \hfill x\left(9x+5\right)+5\left(9x+5\right)\hfill \\ \hfill \left(9x+5\right)\left(x+5\right)\hfill \end{array}$ 
Check. $\begin{array}{}\\ \phantom{\rule{2.5em}{0ex}}\left(9x+5\right)\left(x+5\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}9{x}^{2}+45x+5x+25\hfill \\ \phantom{\rule{2.5em}{0ex}}9{x}^{2}+50x+25\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ 
Try It 7.89
Factor: $16{r}^{2}+30rs+9{s}^{2}$.
Try It 7.90
Factor: $9{u}^{2}+87u+100$.
Remember the very first step in our Strategy for Factoring Polynomials? It was to ask “is there a greatest common factor?” and, if there was, you factor the GCF before going any further. Perfect square trinomials may have a GCF in all three terms and it should be factored out first. And, sometimes, once the GCF has been factored, you will recognize a perfect square trinomial.
Example 7.46
Factor: $36{x}^{2}y48xy+16y$.
Solution
$36{x}^{2}y48xy+16y$  
Is there a GCF? Yes, 4y, so factor it out.  $4y(9{x}^{2}12x+4)$ 
Is this a perfect square trinomial?  
Verify the pattern.  
Factor.  $4y{(3x2)}^{2}$ 
Remember: Keep the factor 4y in the final product.  
Check.  
$4y{(3x2)}^{2}$  
$4y[{\left(3x\right)}^{2}2\xb73x\xb72+{2}^{2}]$  
$4y{\left(9x\right)}^{2}12x+4$  
$36{x}^{2}y48xy+16y\u2713$ 
Try It 7.91
Factor: $8{x}^{2}y24xy+18y$.
Try It 7.92
Factor: $27{p}^{2}q+90pq+75q$.
Factor Differences of Squares
The other special product you saw in the previous chapter was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:
Remember, when you multiply conjugate binomials, the middle terms of the product add to 0. All you have left is a binomial, the difference of squares.
Multiplying conjugates is the only way to get a binomial from the product of two binomials.
Product of Conjugates Pattern
If a and b are real numbers
The product is called a difference of squares.
To factor, we will use the product pattern “in reverse” to factor the difference of squares. A difference of squares factors to a product of conjugates.
Difference of Squares Pattern
If a and b are real numbers,
Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.
Example 7.47
How to Factor Differences of Squares
Factor: ${x}^{2}4$.
Solution
Try It 7.93
Factor: ${h}^{2}81$.
Try It 7.94
Factor: ${k}^{2}121$.
How To
Factor differences of squares.
$\begin{array}{cccc}\mathbf{\text{Step 1.}}\phantom{\rule{0.2em}{0ex}}\text{Does the binomial fit the pattern?}\hfill & & & \hfill {a}^{2}{b}^{2}\hfill \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Is this a difference?}\hfill & & & \hfill \_\_\_\_\_\_\_\_\hfill \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Are the first and last terms perfect squares?}\hfill & & & \\ \mathbf{\text{Step 2.}}\phantom{\rule{0.2em}{0ex}}\text{Write them as squares.}\hfill & & & \hfill {\left(a\right)}^{2}{\left(b\right)}^{2}\hfill \\ \mathbf{\text{Step 3.}}\phantom{\rule{0.2em}{0ex}}\text{Write the product of conjugates.}\hfill & & & \hfill \left(ab\right)\left(a+b\right)\hfill \\ \mathbf{\text{Step 4.}}\phantom{\rule{0.2em}{0ex}}\text{Check by multiplying.}\hfill & & & \end{array}$
It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression ${a}^{2}+{b}^{2}$ is prime!
Don’t forget that 1 is a perfect square. We’ll need to use that fact in the next example.
Example 7.48
Factor: $64{y}^{2}1$.
Solution
Is this a difference? Yes.  
Are the first and last terms perfect squares?  
Yes  write them as squares.  
Factor as the product of conjugates.  
Check by multiplying.  
$(8y1)(8y+1)$  
$64{y}^{2}1\u2713$ 
Try It 7.95
Factor: ${m}^{2}1$.
Try It 7.96
Factor: $81{y}^{2}1$.
Example 7.49
Factor: $121{x}^{2}49{y}^{2}$.
Solution
$\begin{array}{cccc}& & & \hfill 121{x}^{2}49{y}^{2}\hfill \\ \\ \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {\left(11x\right)}^{2}{\left(7y\right)}^{2}\hfill \\ \\ \\ \text{Factor as the product of conjugates.}\hfill & & & \hfill \left(11x7y\right)\left(11x+7y\right)\hfill \\ \\ \\ \text{Check by multiplying.}\hfill & & & \\ \\ \\ \phantom{\rule{2.5em}{0ex}}\left(11x7y\right)\left(11x+7y\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}121{x}^{2}49{y}^{2}\phantom{\rule{0.2em}{0ex}}\u2713\hfill & & & \end{array}$
Try It 7.97
Factor: $196{m}^{2}25{n}^{2}$.
Try It 7.98
Factor: $144{p}^{2}9{q}^{2}$.
The binomial in the next example may look “backwards,” but it’s still the difference of squares.
Example 7.50
Factor: $100{h}^{2}$.
Solution
$100{h}^{2}$  
Is this a difference of squares? Yes.  ${\left(10\right)}^{2}{\left(h\right)}^{2}$ 
Factor as the product of conjugates.  $\left(10h\right)\left(10+h\right)$ 
Check by multiplying. $\begin{array}{}\\ \\ \phantom{\rule{2.5em}{0ex}}\left(10h\right)\left(10+h\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}100{h}^{2}\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ 

Be careful not to rewrite the original expression as ${h}^{2}100$. 
Factor ${h}^{2}100$ on your own and then notice how the result differs from $\left(10h\right)\left(10+h\right)$.
Try It 7.99
Factor: $144{x}^{2}$.
Try It 7.100
Factor: $169{p}^{2}$.
To completely factor the binomial in the next example, we’ll factor a difference of squares twice!
Example 7.51
Factor: ${x}^{4}{y}^{4}$.
Solution
${x}^{4}{y}^{4}$  
Is this a difference of squares? Yes.  ${\left({x}^{2}\right)}^{2}{\left({y}^{2}\right)}^{2}$ 
Factor it as the product of conjugates.  $\left({x}^{2}{y}^{2}\right)\left({x}^{2}+{y}^{2}\right)$ 
Notice the first binomial is also a difference of squares!  $\left({(x)}^{2}{(y)}^{2}\right)\left({x}^{2}+{y}^{2}\right)$ 
Factor it as the product of conjugates. The last factor, the sum of squares, cannot be factored. 
$\left(xy\right)\left(x+y\right)\left({x}^{2}+{y}^{2}\right)$ 
Check by multiplying. $\begin{array}{}\\ \\ \\ \phantom{\rule{2.5em}{0ex}}\left(xy\right)\left(x+y\right)\left({x}^{2}+{y}^{2}\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}\left[\left(xy\right)\left(x+y\right)\right]\left({x}^{2}+{y}^{2}\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}\left({x}^{2}{y}^{2}\right)\left({x}^{2}+{y}^{2}\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}{x}^{4}{y}^{4}\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ 
Try It 7.101
Factor: ${a}^{4}{b}^{4}$.
Try It 7.102
Factor: ${x}^{4}16$.
As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.
Example 7.52
Factor: $8{x}^{2}y98y$.
Solution
$8{x}^{2}y98y$  
Is there a GCF? Yes, 2y—factor it out!  $2y\left(4{x}^{2}49\right)$ 
Is the binomial a difference of squares? Yes.  $2y\left({\left(2x\right)}^{2}{\left(7\right)}^{2}\right)$ 
Factor as a product of conjugates.  $2y\left(2x7\right)\left(2x+7\right)$ 
Check by multiplying. $\begin{array}{}\\ \\ \phantom{\rule{2.5em}{0ex}}2y\left(2x7\right)\left(2x+7\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}2y\left[\left(2x7\right)\left(2x+7\right)\right]\hfill \\ \phantom{\rule{2.5em}{0ex}}2y\left(4{x}^{2}49\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}8{x}^{2}y98y\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ 
Try It 7.103
Factor: $7x{y}^{2}175x$.
Try It 7.104
Factor: $45{a}^{2}b80b$.
Example 7.53
Factor: $6{x}^{2}+96$.
Solution
$6{x}^{2}+96$  
Is there a GCF? Yes, 6—factor it out!  $6\left({x}^{2}+16\right)$ 
Is the binomial a difference of squares? No, it is a sum of squares. Sums of squares do not factor! 

Check by multiplying. $\begin{array}{}\\ \\ \\ \phantom{\rule{2.5em}{0ex}}6\left({x}^{2}+16\right)\hfill \\ \phantom{\rule{2.5em}{0ex}}6{x}^{2}+96\phantom{\rule{0.2em}{0ex}}\u2713\hfill \end{array}$ 
Try It 7.105
Factor: $8{a}^{2}+200$.
Try It 7.106
Factor: $36{y}^{2}+81$.
Factor Sums and Differences of Cubes
There is another special pattern for factoring, one that we did not use when we multiplied polynomials. This is the pattern for the sum and difference of cubes. We will write these formulas first and then check them by multiplication.
We’ll check the first pattern and leave the second to you.
Distribute.  
Multiply.  ${a}^{3}{a}^{2}b+{ab}^{2}+{a}^{2}b{ab}^{2}+{b}^{3}$ 
Combine like terms.  ${a}^{3}+{b}^{3}$ 
Sum and Difference of Cubes Pattern
The two patterns look very similar, don’t they? But notice the signs in the factors. The sign of the binomial factor matches the sign in the original binomial. And the sign of the middle term of the trinomial factor is the opposite of the sign in the original binomial. If you recognize the pattern of the signs, it may help you memorize the patterns.
The trinomial factor in the sum and difference of cubes pattern cannot be factored.
It can be very helpful if you learn to recognize the cubes of the integers from 1 to 10, just like you have learned to recognize squares. We have listed the cubes of the integers from 1 to 10 in Table 7.2.
n  1  2  3  4  5  6  7  8  9  10 

${n}^{3}$  1  8  27  64  125  216  343  512  729  1000 
Example 7.54
How to Factor the Sum or Difference of Cubes
Factor: ${x}^{3}+64$.
Solution
Try It 7.107
Factor: ${x}^{3}+27$.
Try It 7.108
Factor: ${y}^{3}+8$.
How To
Factor the sum or difference of cubes.
To factor the sum or difference of cubes:
 Step 1.
Does the binomial fit the sum or difference of cubes pattern?
 Is it a sum or difference?
 Are the first and last terms perfect cubes?
 Step 2. Write them as cubes.
 Step 3. Use either the sum or difference of cubes pattern.
 Step 4. Simplify inside the parentheses
 Step 5. Check by multiplying the factors.
Example 7.55
Factor: ${x}^{3}1000$.
Solution
This binomial is a difference. The first and last terms are perfect cubes.  
Write the terms as cubes.  
Use the difference of cubes pattern.  
Simplify.  
Check by multiplying.  
Try It 7.109
Factor: ${u}^{3}125$.
Try It 7.110
Factor: ${v}^{3}343$.
Be careful to use the correct signs in the factors of the sum and difference of cubes.
Example 7.56
Factor: $512125{p}^{3}$.
Solution
This binomial is a difference. The first and last terms are perfect cubes.  
Write the terms as cubes.  
Use the difference of cubes pattern.  
Simplify.  
Check by multiplying.  We'll leave the check to you. 
Try It 7.111
Factor: $6427{x}^{3}$.
Try It 7.112
Factor: $278{y}^{3}$.
Example 7.57
Factor: $27{u}^{3}125{v}^{3}$.
Solution
This binomial is a difference. The first and last terms are perfect cubes.  
Write the terms as cubes.  
Use the difference of cubes pattern.  
Simplify.  
Check by multiplying.  We'll leave the check to you. 
Try It 7.113
Factor: $8{x}^{3}27{y}^{3}$.
Try It 7.114
Factor: $1000{m}^{3}125{n}^{3}$.
In the next example, we first factor out the GCF. Then we can recognize the sum of cubes.
Example 7.58
Factor: $5{m}^{3}+40{n}^{3}$.
Solution
Factor the common factor.  
This binomial is a sum. The first and last terms are perfect cubes.  
Write the terms as cubes.  
Use the sum of cubes pattern.  
Simplify. 
Check. To check, you may find it easier to multiply the sum of cubes factors first, then multiply that product by 5. We’ll leave the multiplication for you.
$5\left(\stackrel{}{m+2n}\right)\left(\stackrel{}{{m}^{2}2mn+4{n}^{2}}\right)$
Try It 7.115
Factor: $500{p}^{3}+4{q}^{3}$.
Try It 7.116
Factor: $432{c}^{3}+686{d}^{3}$.
Media
Access these online resources for additional instruction and practice with factoring special products.
Section 7.4 Exercises
Practice Makes Perfect
Factor Perfect Square Trinomials
In the following exercises, factor.
$25{v}^{2}+20v+4$
$49{s}^{2}+154s+121$
$64{z}^{2}16z+1$
$4{p}^{2}52p+169$
$25{r}^{2}60rs+36{s}^{2}$
$100{y}^{2}20y+1$
$100{x}^{2}25x+1$
$64{x}^{2}96x+36$
$90{p}^{3}+300{p}^{2}q+250p{q}^{2}$
Factor Differences of Squares
In the following exercises, factor.
${n}^{2}9$
$169{q}^{2}1$
$49{x}^{2}81{y}^{2}$
$36{p}^{2}49{q}^{2}$
$12125{s}^{2}$
${m}^{4}{n}^{4}$
$98{r}^{3}72r$
$20{b}^{2}+140$
Factor Sums and Differences of Cubes
In the following exercises, factor.
${n}^{3}+512$
${v}^{3}216$
$12527{w}^{3}$
$27{x}^{3}64{y}^{3}$
$6{x}^{3}48{y}^{3}$
$\mathrm{2}{x}^{3}16{y}^{3}$
Mixed Practice
In the following exercises, factor.
$121{x}^{2}144$
$4{p}^{2}100$
$36{y}^{2}+12y+1$
$81{x}^{2}+169$
$27{u}^{3}+1000$
$48{q}^{3}24{q}^{2}+3q$
Everyday Math
Landscaping Sue and Alan are planning to put a 15 foot square swimming pool in their backyard. They will surround the pool with a tiled deck, the same width on all sides. If the width of the deck is w, the total area of the pool and deck is given by the trinomial $4{w}^{2}+60w+225$. Factor the trinomial.
Home repair The height a twelve foot ladder can reach up the side of a building if the ladder’s base is b feet from the building is the square root of the binomial $144{b}^{2}$. Factor the binomial.
Writing Exercises
Why was it important to practice using the binomial squares pattern in the chapter on multiplying polynomials?
How do you recognize the binomial squares pattern?
Maribel factored ${y}^{2}30y+81$ as ${(y9)}^{2}$. Was she right or wrong? How do you know?
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?