## College Algebra 7th Edition

(a) 2$\mathcal{x}$(2$\mathcal{x}$ + 1) (b) 3$\mathcal{x}$$\mathcal{y}$($\mathcal{y}$ - 2$\mathcal{x}$) (c) ($\mathcal{x}$ + 3)($\mathcal{x}$ + 5) (d) ($\mathcal{x}$ - 2)($\mathcal{x}$ + 1) (e) (2$\mathcal{x}$ - 3)($\mathcal{x}$ +4) (f) ($\mathcal{x}$ + 4)($\mathcal{x}$ - 4)
These equations are solved by different types of factoring. The first type of factoring is seen in (a) and (b). The second is seen in c-f. In (a) the problem is: 4$\mathcal{x^2}$ + 2$\mathcal{x}$. To simplify factor 2$\mathcal{x}$ out of the equation. This leaves 2$\mathcal{x}$(2$\mathcal{x}$ + 1). Question (b) is solved using the same method. In (c) the problem is: ($\mathcal{x^2}$ + 8$\mathcal{x}$ +15). When factoring look for numbers that could been multiplied to equal the number with no variable, and added/subtracted to equal the integer with one variable. In this case it is 3 and 5. This leaves ($\mathcal{x}$ + 3)($\mathcal{x}$ + 5). Questions (c), (d), (e), and (f) are solved the same way, with slight differences. In (e) 2$\mathcal{x}$ must be in the same parenthesis with 3 to get the beginning equation. In (f) the integers with $\mathcal{x}$ will cancel out. If you are unsure if you put the integers in the correct places, solve the equation. I feel you get the original it was correct.