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We’ll work one now where the middle term is negative.
Factor: $81{y}^{2}-72y+16$ .
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 ${(a-b)}^{2}$ .
Are the first and last terms perfect squares? | |
Check the middle term. | |
Does is match ${(a-b)}^{2}$ ? Yes. | |
Write the square of a binomial. | |
Check by mulitplying. | |
${(9y-4)}^{2}$ | |
${\left(9y\right)}^{2}-2\cdot 9y\cdot 4+{4}^{2}$ | |
$81{y}^{2}-72y+16\u2713$ |
The next example will be a perfect square trinomial with two variables.
Factor: $36{x}^{2}+84xy+49{y}^{2}$ .
Test each term to verify the pattern. | |
Factor. | |
Check by mulitplying. | |
${(6x+7y)}^{2}$ | |
${\left(6x\right)}^{2}+2\cdot 6x\cdot 7y+{\left(7y\right)}^{2}$ | |
$36{x}^{2}+84xy+49{y}^{2}\u2713$ |
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: $9{x}^{2}+50x+25$ .
$\begin{array}{cccc}& & & \hfill 9{x}^{2}+50x+25\hfill \\ \text{Are the first and last terms perfect squares?}\hfill & & & \hfill {\left(3x\right)}^{2}\phantom{\rule{3em}{0ex}}{\left(5\right)}^{2}\hfill \\ \text{Check the middle term\u2014is it}\phantom{\rule{0.2em}{0ex}}2ab?\hfill & & & \hfill {\left(3x\right)}^{2}{}_{\text{\u2198}}\underset{\underset{30x}{2\left(3x\right)\left(5\right)}}{\text{}}{}_{\text{\u2199}}{\left(5\right)}^{2}\hfill \\ \text{No!}\phantom{\rule{0.2em}{0ex}}30x\ne 50x\hfill & & & \text{This does not fit the pattern!}\hfill \\ \text{Factor using the \u201cac\u201d method.}\hfill & & & \hfill 9{x}^{2}+50x+25\hfill \\ \\ \\ \\ \text{Notice:}\phantom{\rule{0.2em}{0ex}}\begin{array}{c}\hfill ac\hfill \\ \hfill 9\xb725\hfill \\ \hfill 225\hfill \end{array}\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\begin{array}{c}\hfill 5\xb745=225\hfill \\ \hfill 5+45=50\hfill \end{array}\hfill \\ \text{Split the middle term.}\hfill & & & \hfill 9{x}^{2}+5x+45x+25\hfill \\ \text{Factor by grouping.}\hfill & & & \hfill x\left(9x+5\right)+5\left(9x+5\right)\hfill \\ & & & \hfill \left(9x+5\right)\left(x+5\right)\hfill \\ \text{Check.}\hfill & & & \\ \\ \phantom{\rule{2.5em}{0ex}}\left(9x+5\right)\left(x+5\right)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}9{x}^{2}+45x+5x+25\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}9{x}^{2}+50x+25\phantom{\rule{0.2em}{0ex}}\u2713\hfill & & & \end{array}$
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: $36{x}^{2}y-48xy+16y$ .
$36{x}^{2}y-48xy+16y$ | |
Is there a GCF? Yes, 4 y , so factor it out. | $4y(9{x}^{2}-12x+4)$ |
Is this a perfect square trinomial? | |
Verify the pattern. | |
Factor. | $4y{(3x-2)}^{2}$ |
Remember: Keep the factor 4 y in the final product. | |
Check. | |
$4y{(3x-2)}^{2}$ | |
$4y[{\left(3x\right)}^{2}-2\xb73x\xb72+{2}^{2}]$ | |
$4y{\left(9x\right)}^{2}-12x+4$ | |
$36{x}^{2}y-48xy+16y\u2713$ |
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:
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.
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.
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.
Factor: ${k}^{2}-121$ .
$\left(k-11\right)\left(k+11\right)$
$\begin{array}{cccc}\mathbf{\text{Step 1.}}\phantom{\rule{0.2em}{0ex}}\text{Does the binomial fit the pattern?}\hfill & & & \hfill {a}^{2}-{b}^{2}\hfill \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Is this a difference?}\hfill & & & \hfill \_\_\_\_-\_\_\_\_\hfill \\ \phantom{\rule{2.5em}{0ex}}\u2022\phantom{\rule{0.5em}{0ex}}\text{Are the first and last terms perfect squares?}\hfill & & & \\ \mathbf{\text{Step 2.}}\phantom{\rule{0.2em}{0ex}}\text{Write them as squares.}\hfill & & & \hfill {\left(a\right)}^{2}-{\left(b\right)}^{2}\hfill \\ \mathbf{\text{Step 3.}}\phantom{\rule{0.2em}{0ex}}\text{Write the product of conjugates.}\hfill & & & \hfill \left(a-b\right)\left(a+b\right)\hfill \\ \mathbf{\text{Step 4.}}\phantom{\rule{0.2em}{0ex}}\text{Check by multiplying.}\hfill & & & \end{array}$
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