Answer
(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)
Work Step by Step
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.
