How does the AC method of factoring quadratics work?

[krackers]

Lets assume you're given

{ 3x }^{ 2 }+8x-11

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

{ 3x }^{ 2 }-3x+11x-11

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

 

[krackers]

Here's my attempt at trying to prove this.

The general form of a quadratic is

{ Ax }^{ 2 }+Bx+C

When factored, you arrive at

(px+m)(qx+n)

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

1) pq{ (x) }^{ 2 } + pn(x) + qm(x) + mn

2) pq{ (x) }^{ 2 } + (pn+qm)x + mn

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?

 

[Diffy]

Here's my attempt at trying to prove this.

The general form of a quadratic is

{ Ax }^{ 2 }+Bx+C

When factored, you arrive at

(px+m)(qx+n)

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

1) pq{ (x) }^{ 2 } + pn(x) + qm(x) + mn

2) pq{ (x) }^{ 2 } + (pn+qm)x + mn

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

 

[krackers]

Is there a mathematical proof of my last statement though?

 

[AlephZero]

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
$(pq)x^2 + (pn)x +(qm)x + (mn)$
$= px(qx + n) + m(qx + n)$
$= (px + m)(qx + n)$.

 

[Diffy]

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...

 

[krackers]

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.

 

[Studiot]

I think it to be a variation of this method



\begin{array}{l}<br /> {\rm{sum }}\;{\rm{of}}\;{\rm{ roots = - }}\frac{{{\rm{coefficient}}\;{\rm{ of}}\;{\rm{ x}}}}{{{\rm{coefficient }}\;{\rm{of}}\;{\rm{ }}{{\rm{x}}^{\rm{2}}}}} \\ <br /> {\rm{product }}\;{\rm{of}}\;{\rm{ roots = }}\frac{{{\rm{constant}}\;{\rm{ term}}}}{{{\rm{coefficient }}\;{\rm{of }}\;{{\rm{x}}^{\rm{2}}}}} \\ <br /> \end{array}

 

[krackers]

Ooh! That seems to explain it quite well.