Question

Giving Lessons: How To Factor Trinomials using the AC Method

Answer 1

I am giving this lesson, because for the past few days I have seen an increase of factoring questions. If you already know factoring but forget hopefully this will help you, if you already know how to factor then tag someone who doesn’t, and if you don’t know anything about factoring… well you are in the right place! The reason I am making this because I had many problems with this myself, so I don’t want any of you to have trouble.

Lets Get Started! :D

So the first sample question is:

\[x^2 - 6x + 5 = 0\]

Alright so I am going to be using the AC method. First I would like to notify you that the first number in this equation is 1… because if there isn’t a number in front of the 1st variable we will assume that there is a 1 in the front. So.. 1 is a. Now the second part of this equation is

\[6x\]

… so b would be 6. The last part is 5.. so c is 5.

Now we know that in the sample equation 1 is a, 6 is b, and 5 is c. So when using the AC method we have to first multiply a and c. So we have to do 1 times 5. When we do this we would get 5.

So now we will make a list of pairs of numbers that add up to equal b and multiply to equal ac. So we need to find 2 numbers that add up to 6 and multiply to 5. So, take your time thinking

When you are done you should have gotten: 5 and 1. This is because when you add them you get 6 and when you multiply them you get 5.

So we would write that like this:

\[(x - 5)(x - 1)\]

Thats it! Seriously! That is all you have to do using the AC method! So the final answer is

\[(x - 5)(x - 1)\]

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Still don’t get it? Don’t worry! Lets try one more sample question!

So the last sample question is:

\[x^2 + 4x + 3 = 0\]

Alright so we are still going to be using the AC method. Lets identify a, b, and c now. Since there is no number in the front of the equation we are going to assume there is a 1, a is 1. Now the second part of the equation is

\[4x\]

… so b would be 4. The last part is simply 3… so c is 3

Now we know that in the sample equation 1 is a, 4 is b, and 3 is c. So we are now going to multiply a and c like we did in the first problem. So we have to do 1 times 3. When we do this we would get 3.

So now we will make a list of pairs of numbers that add up to equal b and multiply to equal ac. So we need to find 2 numbers that add up to 4 and multiply to 3. So, take your time thinking

When you are done you should have gotten: 3 and 1. This is because when you add them you get 4 and when you multiply you get 3.

So we would write that like this:

\[(x + 3)(x + 1)\]

Thats it! Seriously! That is all you have to do using the AC method! So the final answer is

\[(x + 3)(x + 1)\]

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Answer 2

@UnkleRhaukus Is this right?

Answer 3

Good job!

Answer 4

no
\[x^2 - 6x + 5 = 0\]
\(a=1\), \(b=-6\), \(c=5\)

Answer 5

...
So we need to find 2 numbers that add up to -6 and multiply to 5. So, take your time thinking

When you are done you should have gotten: -5 and -1. This is because when you add them you get -6 and when you multiply them you get 5.