88.4k views
2 votes
Match each polynomial with its factorization.

(I’m about a week behind in classes plz help)

Match each polynomial with its factorization. (I’m about a week behind in classes-example-1

1 Answer

6 votes

Answer:

1 matches to (y + 1)(y - 2)

2 matches to 3(y - 3)(y + 3)

3 matches to (y - 2)(y + 2)(y^2 + 4)

4 matches to (y - 6)^2

5 matches to -4(y + 2)(y - 4)

6 matches to (4y - 3)(y + 2)

Step-by-step explanation:

This is pretty much the same process for every problem, but I'll pick a couple and show you how to handle the different cases.

In #1, you have a straightforward equation y^2 - y - 2, which can possibly have the factors ±1 and ±2. To find the factors of an equation with a coefficient of 1, all you have to do is look at the constant and figure the factors of that constant. With 2, your only options are 1 and 2. What you have to do after that is find some combination of those two numbers so that

(y - ?)(y - ?) = y^2 - y - 2

The easiest way to tackle this one in my opinion is just trial and error. Try plugging in all combinations of -1, 1, -2, and 2 until you hit the jackpot. Here, through trial and error, I found that 2 and -1 work, that is:

(y - 2)(y + 1) = y^2 - y - 2

If you FOIL, or distribute, the left side of the equation, you get the right side.

For one like #2, with a different coefficient (AKA not = 1), you usually have to do some factoring out before you can figure out the roots. 3y^2 - 27 has a common factor of 3, so if you factor 3 out, you have 3(y^2 - 9) left. From there, you just factor (y^2 - 9) and get 3(y - 3)(y + 3). You can do that by noticing y^2 - 9 is a difference of two squares.

#3 once again deals with the difference of two squares, if you imagine y^4 is (y^2)^2 and 16 is 4^2. Factoring that one like any other difference of two squares, you get (y^2 + 4)(y^2 - 4), and the second factor there is yet another difference of two squares, so it can be factored further into (y - 2)(y + 2).

#4 is like example 1. y^2 - 12y + 36 = (y - 6)^2

#5 is like example 2. Factor out the common thing first; the common value in this case is -4, so you get -4(y + 2)(y - 4).

#6 looks really complicated but it isn't so bad. Since you already have the first 5 answers, you can just assume that #6 is the remaining one, but if you want to understand how to solve it: The first thing you want to notice is that you have some like terms that you can combine, -2 and -4, which combine to create -6. Your polynomial becomes 4y^2 + 5y - 6. If you try to factor out anything common, you won't find anything pretty or simple. So when that happens, what I like to do is just take the coefficient of y^2, place it in parentheses like example 1 and then use trial and error.

(4y - ?)(y - ?)

And then I take a look at -6 and I think about the factors of that value. Here, we'd have ±1, ±2, ±3, and ±6 as options. So you take random combinations of those and see what works, or you notice that you need a positive 5 in the middle, so you try combinations that will definitely get you that. Through testing combinations, you'd find (4y - 3)(y + 2), which, if you multiply it out to check, gives you 4y^2 - 3y + 8y - 6, and if you combine the like terms in the middle, you get 4y^2 + 5y - 6, and you're done.

User Xantium
by
6.2k points