Why factor like this?

[Wormman12]

Okay, I really don't know if this is considered algebra, but I had it once during Algebra II so imma wing it and say sorry if it's in the wrong place.

In my e-book for my class, they gave the ratio (x+3/3)=(9/x−3). Once they did the ratio and moved the numbers around, they gave us X^2-36=0. They then factored it to (x-6)(x+6)=0 leaving x to equal 6 or -6. The one thing I don't understand is why they specifically chose to factor by 6. They could have easily have used (x-3)(x+12) or (x-9)(x+4). I know there has to be a special reason they used six, but I am just not seeing or understanding it. I would really like to know and if anyone here is able to answer I would appreciate it.

 

[Deleted member 4993]

Okay, I really don't know if this is considered algebra, but I had it once during Algebra II so imma wing it and say sorry if it's in the wrong place.

In my e-book for my class, they gave the ratio (x+3/3)=(9/x−3). Once they did the ratio and moved the numbers around, they gave us X^2-36=0. They then factored it to (x-6)(x+6)=0 leaving x to equal 6 or -6. The one thing I don't understand is why they specifically chose to factor by 6. They could have easily have used (x-3)(x+12) or (x-9)(x+4). I know there has to be a special reason they used six, but I am just not seeing or understanding it. I would really like to know and if anyone here is able to answer I would appreciate it.

You started with x^2-36=0

Now

(x-3)(x+12) = x^2 + 9x - 36 (not x^2 - 36) and

(x-9)(x+4) = x^2 - 5x - 36 (not x^2 - 36)

That is why!!

 

[MarkFL]

Your two factors of -36 must have a sum of 0, as we could consider that the quadratic is:

\(\displaystyle x^2+0x-36=0\)

The two factors of -36 whose sum is 0 are -6 and 6.

A couple of other ways to view it:

i) \(\displaystyle x^2-36=x^2-6^2\)

Now use the difference of squares formula: \(\displaystyle a^2-b^2=(a+b)(a-b)\).

So, we have:

\(\displaystyle (x+6)(x-6)=0\)

Then use the zero-factor property, and equate each factor to zero and solve for x, to find:

\(\displaystyle x=\pm 6\)

ii) \(\displaystyle x^2-36=0\,\therefore\.x^2=6^2\)

Now use the square root property that if given \(\displaystyle a^2=b^2\) then \(\displaystyle a=\pm b\).

and so we find:

\(\displaystyle x=\pm 6\)

 

[Wormman12]

Your two factors of -36 must have a sum of 0, as we could consider that the quadratic is:

\(\displaystyle x^2+0x-36=0\)

The two factors of -36 whose sum is 0 are -6 and 6.

A couple of other ways to view it:

i) \(\displaystyle x^2-36=x^2-6^2\)

Now use the difference of squares formula: \(\displaystyle a^2-b^2=(a+b)(a-b)\).

So, we have:

\(\displaystyle (x+6)(x-6)=0\)

Then use the zero-factor property, and equate each factor to zero and solve for x, to find:

\(\displaystyle x=\pm 6\)

ii) \(\displaystyle x^2-36=0\,\therefore\.x^2=6^2\)

Now use the square root property that if given \(\displaystyle a^2=b^2\) then \(\displaystyle a=\pm b\).

and so we find:

\(\displaystyle x=\pm 6\)

I think I get it. so the quickest way to find the formula for x^2-n is to find the square root of n and then use the formula (x+sqrt(n))(x-sqrt(n))?

 

[MarkFL]

I would say the quickest way is:

\(\displaystyle x^2-n=0\)

\(\displaystyle x^2=n\)

\(\displaystyle x=\pm\sqrt{n}\)