DVD night

We saw The Lives of Others tonight, which was an intriguing look at the Stasi from the point of view of one of the investigators, with an interesting love triangle between his superior and those who he was spying on.

Solving multi-term quadratic equations

Technique 1

With that 3a²+3b²+6ab-2a-2b-1 formula, I realised that that a representation of the end-result is:

(ua+vb+w)(xa+yb+z) which when expanded out is

uxa²+vy²-(uy+vx)ab-(uz+wx)a-(vz+wy)b+wz

Because wz is -1, we know that one of them is positive and the other is negative. From the -2a and -2b, and given that 3a² is from 1*3 and -1 is from, 1*1, we can tell that -2a is achieved from -3a+a. So we’ll put the negative on the left and the large terms on the right.

(ua+vb-1)(xa+yb+1)

The confirmation comes from (vz+wy)b from which we get that (v-y)=-2, or that v=y-2. We know from 3b² that v and y multiply to 3, so we now know that y is 3 and v is 1.

(ua+b-1)(xa+3b+1)

Doing similar with (uz+wx)a gives us (u-x)=-2 and that u=x-2 which from 3a² leads us to x being 3 and u being 1

(a+b-1)(3a+3b+1)

Which when expanded out is

3a² + 3b² + (3ab+3ab) + (-3+1)a + (-3+1)b – 1

or

3a² + 3b² + 6ab – 2a – 2b – 1

Which is what we began with from the beginning.

Technique 2

There are other ways though.

We can find the root factors of 3a², 3b², and -1, and from (1*3, 1*3, -1*1) resulting in 1 and 3 we can determine that +6ab, -2a and -2b must eventuate from +3+3 and -3+1

So we know that a and b have 1 and 3 multipliers.

Starting with (?a+?b-1)(?a+?b+1)

we can fill in the a terms (1a+?b-1)(3a+?b+1)

and then the b terms can be filled in (1a+1b-1)(3a+3b+1)

To arrive at the solution.

Technique 3

And I’ve just realised. That they can be solved separately too.

We can pull out the b terms to get 3a²-2a-1 which easily factors to (a-1)(3a+1)

and we can pull out the a terms to get 3b²-2b-1 which factors to (b-1)(3b+1)

So the solution is now easily found to be (a+b-1)(3a+3b+1)

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