7.4 Factor special products  (Page 3/7)

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.
$(8y-1)(8y+1)$
$64y^2-1✓$

Solution

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

Solution

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

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

Solution

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

Solution

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

Solution

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