# 7.4 Factor special products  (Page 2/7)

 Page 2 / 7

We’ll work one now where the middle term is negative.

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 ${\left(a-b\right)}^{2}$ .

 Are the first and last terms perfect squares? Check the middle term. Does is match ${\left(a-b\right)}^{2}$ ? Yes. Write the square of a binomial. Check by mulitplying. ${\left(9y-4\right)}^{2}$ ${\left(9y\right)}^{2}-2\cdot 9y\cdot 4+{4}^{2}$ $81{y}^{2}-72y+16✓$

Factor: $64{y}^{2}-80y+25$ .

${\left(8y-5\right)}^{2}$

Factor: $16{z}^{2}-72z+81$ .

${\left(4z-9\right)}^{2}$

The next example will be a perfect square trinomial with two variables.

Factor: $36{x}^{2}+84xy+49{y}^{2}$ .

## Solution

 Test each term to verify the pattern. Factor. Check by mulitplying. ${\left(6x+7y\right)}^{2}$ ${\left(6x\right)}^{2}+2\cdot 6x\cdot 7y+{\left(7y\right)}^{2}$ $36{x}^{2}+84xy+49{y}^{2}✓$

Factor: $49{x}^{2}+84xy+36{y}^{2}$ .

${\left(7x+6y\right)}^{2}$

Factor: $64{m}^{2}+112mn+49{n}^{2}$ .

${\left(8m+7n\right)}^{2}$

Factor: $9{x}^{2}+50x+25$ .

## Solution

$\begin{array}{cccc}& & & \hfill 9{x}^{2}+50x+25\hfill \\ \text{Are the first and last terms perfect squares?}\hfill & & & \hfill {\left(3x\right)}^{2}\phantom{\rule{3em}{0ex}}{\left(5\right)}^{2}\hfill \\ \text{Check the middle term—is it}\phantom{\rule{0.2em}{0ex}}2ab?\hfill & & & \hfill {\left(3x\right)}^{2}{}_{\text{↘}}\underset{\underset{30x}{2\left(3x\right)\left(5\right)}}{\text{}}{}_{\text{↙}}{\left(5\right)}^{2}\hfill \\ \text{No!}\phantom{\rule{0.2em}{0ex}}30x\ne 50x\hfill & & & \text{This does not fit the pattern!}\hfill \\ \text{Factor using the “ac” method.}\hfill & & & \hfill 9{x}^{2}+50x+25\hfill \\ \\ \\ \\ \text{Notice:}\phantom{\rule{0.2em}{0ex}}\begin{array}{c}\hfill ac\hfill \\ \hfill 9·25\hfill \\ \hfill 225\hfill \end{array}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\begin{array}{c}\hfill 5·45=225\hfill \\ \hfill 5+45=50\hfill \end{array}\hfill \\ \text{Split the middle term.}\hfill & & & \hfill 9{x}^{2}+5x+45x+25\hfill \\ \text{Factor by grouping.}\hfill & & & \hfill x\left(9x+5\right)+5\left(9x+5\right)\hfill \\ & & & \hfill \left(9x+5\right)\left(x+5\right)\hfill \\ \text{Check.}\hfill & & & \\ \\ \phantom{\rule{2.5em}{0ex}}\left(9x+5\right)\left(x+5\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}9{x}^{2}+45x+5x+25\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}9{x}^{2}+50x+25\phantom{\rule{0.2em}{0ex}}✓\hfill & & & \end{array}$

Factor: $16{r}^{2}+30rs+9{s}^{2}$ .

$\left(8r+3s\right)\left(2r+3s\right)$

Factor: $9{u}^{2}+87u+100$ .

$\left(3u+4\right)\left(3u+25\right)$

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.

Factor: $36{x}^{2}y-48xy+16y$ .

## Solution

 $36{x}^{2}y-48xy+16y$ Is there a GCF? Yes, 4 y , so factor it out. $4y\left(9{x}^{2}-12x+4\right)$ Is this a perfect square trinomial? Verify the pattern. Factor. $4y{\left(3x-2\right)}^{2}$ Remember: Keep the factor 4 y in the final product. Check. $4y{\left(3x-2\right)}^{2}$ $4y\left[{\left(3x\right)}^{2}-2·3x·2+{2}^{2}\right]$ $4y{\left(9x\right)}^{2}-12x+4$ $36{x}^{2}y-48xy+16y✓$

Factor: $8{x}^{2}y-24xy+18y$ .

$2y{\left(2x-3\right)}^{2}$

Factor: $27{p}^{2}q+90pq+75q$ .

$3q{\left(3p+5\right)}^{2}$

## Factor differences of squares

The other special product you saw in the previous was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

$\begin{array}{c}\hfill \left(3x-4\right)\left(3x+4\right)\hfill \\ \hfill 9{x}^{2}-16\hfill \end{array}$

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

$\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$

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.

## How to factor differences of squares

Factor: ${x}^{2}-4$ .

## Solution

Factor: ${h}^{2}-81$ .

$\left(h-9\right)\left(h+9\right)$

Factor: ${k}^{2}-121$ .

$\left(k-11\right)\left(k+11\right)$

## Factor differences of squares.

$\begin{array}{cccc}\mathbf{\text{Step 1.}}\phantom{\rule{0.2em}{0ex}}\text{Does the binomial fit the pattern?}\hfill & & & \hfill {a}^{2}-{b}^{2}\hfill \\ \phantom{\rule{2.5em}{0ex}}•\phantom{\rule{0.5em}{0ex}}\text{Is this a difference?}\hfill & & & \hfill ____-____\hfill \\ \phantom{\rule{2.5em}{0ex}}•\phantom{\rule{0.5em}{0ex}}\text{Are the first and last terms perfect squares?}\hfill & & & \\ \mathbf{\text{Step 2.}}\phantom{\rule{0.2em}{0ex}}\text{Write them as squares.}\hfill & & & \hfill {\left(a\right)}^{2}-{\left(b\right)}^{2}\hfill \\ \mathbf{\text{Step 3.}}\phantom{\rule{0.2em}{0ex}}\text{Write the product of conjugates.}\hfill & & & \hfill \left(a-b\right)\left(a+b\right)\hfill \\ \mathbf{\text{Step 4.}}\phantom{\rule{0.2em}{0ex}}\text{Check by multiplying.}\hfill & & & \end{array}$

Bruce drives his car for his job. The equation R=0.575m+42 models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day. Find the amount Bruce is reimbursed on a day when he drives 220 miles.
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