7.4 Factor Special Products

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.
${(9y-4)}^2$
${\left(9y\right)}^2-2\cdot 9y\cdot 4+4^2$
$81y^2-72y+16✓$

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

$\begin{array}{cccc}& & & \hfill 121x^2-49y^2\hfill \\ \\ \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {\left(11x\right)}^2-{\left(7y\right)}^2\hfill \\ \\ \\ \text{Factor as the product of conjugates.}\hfill & & & \hfill \left(11x-7y\right)\left(11x+7y\right)\hfill \\ \\ \\ \text{Check by multiplying.}\hfill & & & \\ \\ \\ \left(11x-7y\right)\left(11x+7y\right)\hfill & & & \\ 121x^2-49y^2✓\hfill & & & \end{array}$

$\begin{array}{cccc}& & & \hfill 100-h^2\hfill \\ \\ \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {\left(10\right)}^2-{\left(h\right)}^2\hfill \\ \\ \\ \text{Factor as the product of conjugates.}\hfill & & & \hfill \left(10-h\right)\left(10+h\right)\hfill \\ \\ \\ \text{Check by multiplying.}\hfill & & & \\ \\ \\ \left(10-h\right)\left(10+h\right)\hfill & & & \\ 100-h^2✓\hfill & & & \end{array}$

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 $\left(10-h\right)\left(10+h\right)$.

$\begin{array}{cccc}& & & \hfill x^4-y^4\hfill \\ \\ \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {\left(x^2\right)}^2-{\left(y^2\right)}^2\hfill \\ \\ \\ \text{Factor it as the product of conjugates.}\hfill & & & \hfill \left(x^2-y^2\right)\left(x^2+y^2\right)\hfill \\ \\ \\ \text{Notice the first binomial is also a difference of squares!}\hfill & & & \hfill \left({(x)}^2-{(y)}^2\right)\left(x^2+y^2\right)\hfill \\ \\ \\ \text{Factor it as the product of conjugates. The last}\hfill & & & \hfill \left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\hfill \\ \text{factor, the sum of squares, cannot be factored.}\hfill & & & \\ \\ \\ \text{Check by multiplying.}\hfill & & & \\ \\ \\ \left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\hfill & & & \\ \left[\left(x-y\right)\left(x+y\right)\right]\left(x^2+y^2\right)\hfill & & & \\ \left(x^2-y^2\right)\left(x^2+y^2\right)\hfill & & & \\ x^4-y^4✓\hfill & & & \end{array}$

$\begin{array}{cccc}& & & \hfill 8x^{2}y-98y\hfill \\ \\ \\ \text{Is there a GCF? Yes,}2y\text{—factor it out!}\hfill & & & \hfill 2y\left(4x^2-49\right)\hfill \\ \\ \\ \text{Is the binomial a difference of squares? Yes.}\hfill & & & \hfill 2y\left({\left(2x\right)}^2-{\left(7\right)}^2\right)\hfill \\ \\ \\ \text{Factor as a product of conjugates.}\hfill & & & \hfill 2y\left(2x-7\right)\left(2x+7\right)\hfill \\ \\ \\ \text{Check by multiplying.}\hfill & & & \\ \\ \\ 2y\left(2x-7\right)\left(2x+7\right)\hfill & & & \\ 2y\left[\left(2x-7\right)\left(2x+7\right)\right]\hfill & & & \\ 2y\left(4x^2-49\right)\hfill & & & \\ 8x^{2}y-98y✓\hfill & & & \end{array}$

$\begin{array}{cccc}& & & \hfill 6x^2+96\hfill \\ \\ \\ \text{Is there a GCF? Yes, 6—factor it out!}\hfill & & & \hfill 6\left(x^2+16\right)\hfill \\ \\ \\ \text{Is the binomial a difference of squares? No, it}\hfill & & & \\ \text{is a sum of squares. Sums of squares do not factor!}\hfill & & & \\ \\ \\ \text{Check by multiplying.}\hfill & & & \\ \\ \\ 6\left(x^2+16\right)\hfill & & & \\ 6x^2+96✓\hfill & & & \end{array}$

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.

$5\left(\stackrel{}{m+2n}\right)\left(\stackrel{}{m^2-2mn+4n^2}\right)$