Depending on what you see when you approach this problem, you may choose to solve by distribution or factoring. You may notice that there is an f in each term of the expression. In this case, you may choose to factor the expression tof(gf​+1), but that doesn't give you a possible answer; however, you can eliminate (B), since f(gf​+f) is not equivalent to your expression. Now you are left with an expression you may find hard to manipulate. Go back and look at the mathematics behind the initial factorization so the next step in the process will make more sense.When you factor an f out of the expression, what you are really doing is this: Multiply the expression by 1 so that you do not change the expression.ff​(gf2​+f)f×f1​(gf2​+f)Distribute the f1​f(gf2​×f1​+f×f1​)Cancel where possible.f(gf​+1)You can follow the exact same steps to factor an gf​ like you see in (A).gf​gf​​(gf2​+f)gf​×fg​(gf2​+f)gf​(gf2​×fg​+f×fg​)gf​(f+g)This leaves you with the same expression as (A).If you notice that the answers are all expressions themselves, you may choose to distribute the variable in front of the parentheses instead of trying to factor the expression.Start with (A): gf​(f+g)Distribute to each term within the binomial: f×gf​+g×gf​Cancel where you can: gf2​+fBoth methods give you the same answer; however, this type of algebra leaves you open to making mistakes. In the next chapter, you will discover a third way to approach this problem that you may find even easier than the previous methods.