How does the AC method of factoring quadratics work?

In summary: The sum of the roots is the coefficient of x divided by the coefficient of x squared, and the product of the roots is the constant term divided by the coefficient of x squared. So when you multiply the coefficients of x squared and the constant, you get AC. And then finding the factors of AC that add up to B would essentially be finding the roots that satisfy those conditions. Is that correct?In summary, the AC method for factoring a quadratic involves multiplying the coefficients of x squared and the constant term to get AC, and then finding the factors of AC that add up to the coefficient of x. This allows you to split the coefficient of x into two separate addends, which can then be factored into the final form of (
  • #1
krackers
72
0
Lets assume you're given

[itex]{ 3x }^{ 2 }+8x-11[/itex]

And you want to factor it. With the AC method you multiple 3 and -11 giving you -33. Then you find the factors of -33 that add up to 8. 11 and -3, in this case. Then you rewrite the quadratic as

[itex]{ 3x }^{ 2 }-3x+11x-11[/itex]

From there, you factor each part independently giving:

3x(x-1)+11(x-1)

And finally, factor out (x-1) to get:

(3x+11)(x-1).

However, I do not have any understanding as to how this works.

Thanks
 
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  • #2
Here's my attempt at trying to prove this.

The general form of a quadratic is

[itex]{ Ax }^{ 2 }+Bx+C[/itex]

When factored, you arrive at

(px+m)(qx+n)

If you work backwards and distribute the factored form you get:

1) [itex]pq{ (x) }^{ 2 } + pn(x) + qm(x) + mn[/itex]

2) [itex]pq{ (x) }^{ 2 } + (pn+qm)x + mn[/itex]

You know that if you get back to step 1 then you can successfully factor. However, to do that, you need to split the combined sum of pn + qm back into the two separate addends.

In the AC method, you multiple A and C and find the factors of the product adding up to B.
In this case, you would multiply pq and mn getting pqmn.

So you would find the factors of pqmn adding up (pn + qm). As a result, you would need to get pn and qm.

Intuitively, it makes sense, but is there a mathematical proof for this?
 
  • #3
krackers said:
Here's my attempt at trying to prove this.

The general form of a quadratic is

[itex]{ Ax }^{ 2 }+Bx+C[/itex]

When factored, you arrive at

(px+m)(qx+n)

If you work backwards and distribute the factored form you get:

1) [itex]pq{ (x) }^{ 2 } + pn(x) + qm(x) + mn[/itex]

2) [itex]pq{ (x) }^{ 2 } + (pn+qm)x + mn[/itex]

You know that if you get back to step 1 then you can successfully factor. However, to do that, you need to split the combined sum of pn + qm back into the two separate addends.

In the AC method, you multiple A and C and find the factors of the product adding up to B.
In this case, you would multiply pq and mn getting pqmn.

So you would find the factors of pqmn adding up (pn + qm). As a result, you would need to get pn and qm.

Intuitively, it makes sense, but is there a mathematical proof for this?

I think you just gave one :smile:
 
  • #4
Is there a mathematical proof of my last statement though?
 
  • #5
krackers said:
Intuitively, it makes sense, but is there a mathematical proof for this?

You almost proved it yourself.
You want ##(pq)x^2 + (pn+qm)x + (mn) = Ax^2 + Bx + C## for every possible value of x. The only way to do that is when the coefficients of each power of x are the same.
In other words
##pq = A##,
##pn+qm = B##, and
##mn = C##.

The reason it works is because ##(pq)(mn) = AC## and also ##(pn)(qm) = AC##. So you find two factors of AC that add up to B, and then
##(pq)x^2 + (pn)x +(qm)x + (mn)##
## = px(qx + n) + m(qx + n)##
##= (px + m)(qx + n)##.
 
  • #6
krackers said:
Is there a mathematical proof of my last statement though?

I am not sure what is left to show?

A*C = pqmn.
B = pn+qm
pq*x^2 + pn*x + qm*x + mn

= (pq*x^2 + pn*x) + (qm*x + mn)

= p*x(q*x + n) + m*(q*x + n)

Now factor the q*x + n and get

= (p*x + m)(q*x + n)Edit:

I know remember this was tough for me to see when I first saw it.

Let (q*x + n) = Z

then p*x(q*x + n) + m*(q*x + n)

becomes

p*x*Z + m*Z

Now it is easier to see why you can factor because the above equation becomes

Z*(p*x + m)

Now substitute back in (q*x + n) for Z...
 
  • #7
Diffy said:
I am not sure what is left to show?

A*C = pqmn.
B = pn+qm



pq*x^2 + pn*x + qm*x + mn

= (pq*x^2 + pn*x) + (qm*x + mn)

= p*x(q*x + n) + m*(q*x + n)

Now factor the q*x + n and get

= (p*x + m)(q*x + n)


Edit:

I know remember this was tough for me to see when I first saw it.

Let (q*x + n) = Z

then p*x(q*x + n) + m*(q*x + n)

becomes

p*x*Z + m*Z

Now it is easier to see why you can factor because the above equation becomes

Z*(p*x + m)

Now substitute back in (q*x + n) for Z...

I already knew that... I was looking for the reason you multiply A with C and find the factors of that adding up to B. In essence, how multiplying AC and finding the factors adding up to B allows you to split B into pn and qm. I know how to factor from there.
 
  • #8
I think it to be a variation of this method



[tex]\begin{array}{l}
{\rm{sum }}\;{\rm{of}}\;{\rm{ roots = - }}\frac{{{\rm{coefficient}}\;{\rm{ of}}\;{\rm{ x}}}}{{{\rm{coefficient }}\;{\rm{of}}\;{\rm{ }}{{\rm{x}}^{\rm{2}}}}} \\
{\rm{product }}\;{\rm{of}}\;{\rm{ roots = }}\frac{{{\rm{constant}}\;{\rm{ term}}}}{{{\rm{coefficient }}\;{\rm{of }}\;{{\rm{x}}^{\rm{2}}}}} \\
\end{array}[/tex]
 
  • #9
Ooh! That seems to explain it quite well.
 

1. How does the AC method of factoring quadratics work?

The AC method of factoring quadratics is a technique used to factor quadratic equations with the form ax^2 + bx + c. It involves finding two numbers, a and c, whose product is equal to ac and whose sum is equal to b. These numbers are then used to rewrite the quadratic equation as (ax^2 + ax) + (cx + c). The common factor, a, is factored out from the first two terms and the common factor, c, is factored out from the last two terms. This results in the factored form of the quadratic equation, (ax + c)(x + c).

2. What is the purpose of using the AC method in factoring quadratics?

The AC method is used in factoring quadratics to simplify the process of factoring and to make it more efficient. It can be particularly helpful when the coefficient of x^2 is not equal to 1, as it allows for quicker identification of the factors.

3. Can the AC method be used for all types of quadratic equations?

Yes, the AC method can be used for all types of quadratic equations, including those with a leading coefficient that is not equal to 1. However, it may not always result in a factorable quadratic, in which case other methods can be used.

4. Are there any limitations to the AC method of factoring quadratics?

The AC method may not work for all quadratic equations, particularly those with complex or irrational solutions. It also relies on the coefficients a, b, and c being integers, so it may not be applicable in all cases.

5. How does the AC method differ from other factoring methods?

The AC method differs from other factoring methods, such as the grouping method or the quadratic formula, in that it specifically focuses on finding two numbers whose product is equal to ac and whose sum is equal to b. This method is also only applicable to quadratic equations, whereas other methods can be used for factoring polynomials of higher degrees.

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