How to factor faster with the X method?

[Guest]

Hello,

Well, I was just wondering is a trick to getting the two numbers whose sum and multiplication is one number and another faster when working with big numbers using the X shape method. For example, in this problem:

I make the X shape for it:

But finding what the neumbers are on the left and right side of the X takes me forever to figure out. I tried randomly choosing numbers and trying them out but that just takes way to long. Is there a faster way to figure out those two numbers by hand without a calculator or computer? This isn't just for this problem.

Thanks[/img]

 

[stapel]

I've never heard of the "X method", and it doesn't seem to be particularly helpful.

When dealing with big numbers, just grab your calculator and see what you can get. You multiplied 25 = 5×5 and 49 = 7×7, so you know that the only divisors of 25×49 = 1225 are going to be multiples of 5 and/or 7.

On the off chance that this is a perfect square (since 25x² = (5x)² and 49 = 7²), check that first:

. . . . .(2)(5x)(7) = 70x

There ya go: it's a perfect-square trinomial, and you're done:

. . . . .(5x + 7)²

But, in general, just plug-n-chug, checking for (in this case) sums:

. . . . .1, 1225: sum of 1226 - NO
. . . . .5, 245: sum of 250 - NO
. . . . .7, 175: sum of 182 - NO
. . . . .25, 49: sum of 74 - NO
. . . . .35, 35: sum of 70 - YES

Then factor as usual.

Eliz.

 

[soroban]

Hello, AirForceOne!

Yes, there is a method.
It's rather primitive but it's much faster than random guessing.


Example: Factor \(\displaystyle 24x^2\,-\,23x\,-\,12\)

Multiply: \(\displaystyle \,24\,\times\,12\:=\:288\)

Because of the final minus-sign, we want to factor 288
. . onto two parts whose difference is 23.


Divide 288 by 1,2,3,4, ... and note which divisions "come out even".

For example: \(\displaystyle \,288\,\div\,1\:=\:288\) . . . We have a pair of factors: \(\displaystyle (1,288)\)

\(\displaystyle 288\,\div\,1\:=\:288\;\;(1,288)\)
\(\displaystyle 288\,\div\,2\:=\:144\;\2,144)\)
\(\displaystyle 288\,\div\,3\:=\:96\;\;\;\;(3,96)\)
\(\displaystyle 288\,\div\,4\:=\:72\;\;\;\4,72)\)
\(\displaystyle 288\,\div\,5\:=\:?\)
\(\displaystyle 288\,\div\,6\:=\:48\;\;\;\;(6,48)\)
\(\displaystyle 288\,\div\,7\:=\:?\)
\(\displaystyle 288\,\div\,8\:=\:36\;\;\;\;(8,36)\)
\(\displaystyle 288\,\div\,9\:=\:32\;\;\;\;(9,32)\)
\(\displaystyle 288\,\div\,10\:=\:?\)
\(\displaystyle 288\,\div\,11\:=\:?\)
\(\displaystyle 288\,\div\,12\:=\:24\;\;(12,24)\)
\(\displaystyle 288\,\div\,13\:=\:?\)
\(\displaystyle 288\,\div\,14\:=\:?\)
\(\displaystyle 288\,\div\,15\:=\:?\)
\(\displaystyle 288\,\div\,16\:=\:18\;\;(16,18)\)
\(\displaystyle 288\,\div\,17\:=\:?\;\;\;\;\;\uparrow\)
\(\displaystyle 288\,\div\,18\:=\:16\;\;(18,16)\)

Since \(\displaystyle (18,16)\,=\,(16,18)\), we can stop ... we've reached the "cross-over" point.

We have now found all the two-number factorings of 288.
We wanted the pair that has a difference of 23 ... remember?
. . That pair is: \(\displaystyle \,(9,32)\)

Since the middle number is \(\displaystyle \,-23x\), we will use \(\displaystyle +9\) and \(\displaystyle \,-32.\)


Now we can complete the factoring:

. . . . . . . . \(\displaystyle 24x^2\,-\,23x\,-\,12\)
. . . . . . \(\displaystyle 24x^2\,+\,\overbrace{9x\,-\,32x}\,-\,12\)

Factor: \(\displaystyle \,3x(8x\,+\,3)\,-\,4(8x\,+\,3)\)

Factor: \(\displaystyle \,(8x\,+\,3)(3x\,-\,4)\;\;\) . . . There!

 

[Guest]

Soroban and Stapel, thanks a ton!