Factorization

[hazy tendon]

How do you solve these equations?
2x^2 - 6x + 4

2x^2 + 4x

3xy -6y

14xy^2 + 21x^2y

20P - 5k

[stiff spade]

those aren't equations

how do you factor those?

well for some of them you can look for all the terms having 1 common factor

and then un-distribute that

for example

10a + 5b => 5(2a + b)

[hazy tendon]

@stiff spade Can you do number 1 as an example

[stiff spade]

that one you need to factor using quadratic methods

one way to do that is to factor out the leading coefficient:

2(x^2 - 3x + 2)

then find the 0's of the quadratic inside:
(call them r,s)

then finally you factor it as 2(x-r)(x-s)

[hazy tendon]

But my teacher said 2 (x^2 -3x +2) was wrong,

and (2x-4) (x-1) was the answer

[stiff spade]

ik that wasnt the last step

[hazy tendon]

Then what's the next step?

[stiff spade]

find the 0's

in this case they are 1 and 2

you can do that with a graph or with quadratic formula

at some point you get good enough at factoring to then solve quadratics with it

[hazy tendon]

But he didn't use the quadratic formula

[stiff spade]

what did he do

[hazy tendon]

He said

-2 x -4 = 8 so

(2x^2x) (-4x +4)

2x (x-1) -4 (x-1)

(2x - 4) (x-1)

[stiff spade]

ummmmm

i do not get this logic at all

[hazy tendon]

Well you see...

He put each under a, b and c

2 =a
b = -6
4 = c

[stiff spade]

b should be -6

[hazy tendon]

Yeah then he did ac to get 8

He then found two factors of 8, then when multiplied = 8 and when added = -6

[stiff spade]

yes

-4 and -2

that won't always work, though

[hazy tendon]

yes

[stiff spade]

so what question do you have

[hazy tendon]

The ones above that I showed you

[stiff spade]

like 2x^2 + 4x?

[hazy tendon]

Also I have to learn his method because they'll probably give that in class. If there are any others I can learn that later

no

2x ^2 -6x + 4

[stiff spade]

you just did it

you just explained how the teacher did it

[hazy tendon]

Yes but I don't know how to do it for the other equations

The rest don't have c

[stiff spade]

true

when there's only 2 terms the way you would factor is by looking for something they are all divisible by

for example 2x^2 + 4x they are both divisible by 2x

so you change it to 2x(x + 2)

[hazy tendon]

That's just the normal factorization method I did, but I got it wrong :c

NOOOOOOOOOOOOOOOOOOO!!!!!!! WHAT SHALL I EVER DO! :C

I must follow your method z

It's you or nothing...

@stiff spade Thank you +1 💝

[glossy night]

@hazy tendon has given 1 rep to @stiff spade

[hazy tendon]

@summer cloak Do you possibly know the method my teacher used, and why not the one which z used? as well as if his method can apply to the other questions?

[stiff spade]

the method i said in general only works for 2 term things

[hazy tendon]

@stiff spade I see, but My OTHER math teacher did it with another equation that was like xy +ay + bx + cx and did it your way

[stiff spade]

axy + bx + cy + d?

[hazy tendon]

actually it was

ax + by + bx + ay

the answer was x (a+b) + y (a+b)

[stiff spade]

ah

(x + y)(a + b)

[hazy tendon]

yeah

[odd isle]

Alright, so what do you need help with?

[hazy tendon]

I need to learn how to incorporate the method my teacher showed me into the other equations he gave

The first question was 2x^2 -6x + 4.

To solve it he did:
-2 x -4 = 8 so

(2x^2x) (-4x +4)

2x (x-1) -4 (x-1)

(2x - 4) (x-1)

@odd isle It basically requires me to multiply ac, the problem is for the other equations there is only a and b. So what do you do from there?

[odd isle]

Hm. Not sure if I've heard of this method.

[hazy tendon]

Ikr 🤣

[odd isle]

I would do it like this:
2x^2 - 6x + 4 = 2(x^2 - 3x + 2)
Then we invoke Vieta to see that this factors as follows:
2(x^2 - 3x + 2) = 2(x - 1)(x - 2)

[hazy tendon]

I got this at first, 2(x^2 - 3x + 2) but... he said that was wrong. How do you turn it into 2 (x-1) (x-2)?

[odd isle]

Well, that's not the same algorithm, but it isn't inherently wrong.
As for the factoring, Vieta's formulas give:
x1 + x2 = 3
x1 x2 = 2
From here it's easy to guess x1 = 1, x2 = 2. So, x^2 - 3x + 2 = (x - 1)(x - 2).

[hazy tendon]

[odd isle]

What?

[hazy tendon]

I'm a bit confused. Can you say why it may not be viable, or why it isn't completely right?

[stiff spade]

what?

that was right

2(x^2 - 3x + 2) isn't right because you can factor further

[hazy tendon]

Hi 😄 👋

[odd isle]

Or the teacher just wanted a different algorithm.

[hazy tendon]

I see

[odd isle]

Still a pretty jerk move to call this wrong, but oh well.

[stiff spade]

to call this wrong?

[odd isle]

I mean, sure, that's not the answer yet, but it's the correct way to the answer.

[hazy tendon]

well he just showed me the other way, he didn't say it was wrong outright

[odd isle]

Oh.

Well, I'm not familiar with that approach, sorry.

[hazy tendon]

no problem, what was the other method you used?

[peak fern]

...wait, is the problem the method?

[hazy tendon]

Indeed

[peak fern]

Like, what is the actual final answer you provided?

[odd isle]

I know three basic approaches for solving a quadratic equation:

1. Using the quadratic formula.
2. Completing the square.
3. Using Vieta's formulas if the coefficients are integers - this is what I used.

[hazy tendon]

no, just how to incorporate his method into other equations.

[hazy tendon]

Which one is the fastest?

Quadratic formula 💀

[stiff spade]

if it's factorable, vietas will be fastest

[odd isle]

Yeah.

[stiff spade]

well if the numbers are small

[odd isle]

But Vieta might not help.

In that case quadratic formula is faster.

[stiff spade]

(x - phi)(x + 1/phi)

[odd isle]

But sometimes completing the square is a bit faster.

Depends on the case, really.

[stiff spade]

ye

if you see x^2 + [even number]x = integer, then maybe completing square is fastest

[odd isle]

Yeah.

[hazy tendon]

I'm aware with completing the square but, would you get (x-1) (x-2) with it? Usually using that method you get something like (x-1) +2 you don't usually get 2 brackets from what I was taught.

[stiff spade]

hm?

here is how we would complete the square here

x^2 - 3x + 2 = 0

x^2 - 3x = -2

x^2 - 3x + 9/4 = 9/4 - 2
x^2 - 3x + 9/4 = 1/4

(x - 3/2)^2 = 1/4

(x - 3/2) = +- 1/2

anyway you can see it's pretty much a lot of effort

[hazy tendon]

I was talking about this

[stiff spade]

i really do not get this

[odd isle]

Or, if you want to factor, use the difference of squares formula:
(x - 3/2)^2 - 1/4 = 0
(x - 3/2)^2 - (1/2)^2 = 0
(x - 3/2 - 1/2)(x - 3/2 + 1/2) = 0
(x - 2)(x - 1) = 0

Yeah, same.

[stiff spade]

ooooohh yessss

difference of squares

i have not thought of that before

[odd isle]

Why learn this approach at all? The three listed above work absolutely fine.

[hazy tendon]

Difference of squares? 🤨

[odd isle]

a^2 - b^2 = (a - b)(a + b)

[stiff spade]

well this is incomprehensible

[odd isle]

Yeah, I agree.

Did your teacher provide the general algorithm?

[hazy tendon]

Only the example he gave

[odd isle]

Because these three are easily explainable generally.

[hazy tendon]

Why do teachers ALWAYS give homework that's completely different from the original example they give

[stiff spade]

hm

[odd isle]

So, if your teacher's algorithm is viable in any way, it should also be able to be explained generally.

[stiff spade]

if it's factoring it's probably not completely different

[odd isle]

Wait, really? This is the only example that was given for this method, not even a general explanation? Very weird...

[stiff spade]

the state of education

[hazy tendon]

I mean it was revision for extra classes. I came late due to training for examination. But no one there knew either

[odd isle]

Contact your teacher and ask him to explain it again.

[hazy tendon]

I understand what he did, but I don't understand how to apply it to equations that don't have a third coefficient

[odd isle]

Well, ask that, then.

[hazy tendon]

Yes ofcourse...

[odd isle]

About the general case of ax^2 + bx + c, I mean.

[hazy tendon]

Indeed

I'll learn your method as well

[odd isle]

Wait, you mean you haven't learned any other methods yet?

Very odd, to give an unknown method first...

[hazy tendon]

I've learned completing the square

[odd isle]

Ah, ok.

That's good.

[hazy tendon]

and factorizing with grouping 🙂

[odd isle]

Well... Grouping isn't really a general method. It works sometimes, but not always.

[hazy tendon]

I see

Wait but don't you need to find the square root when factorizing the difference of 2 squares? What would you do for 2x^2 + 4x and 3xy - 6y... and other equations that don't have a whole number for the square root

[odd isle]

Well, quadratic equations in general can have non-integer solutions.

x^2 - 2 = 0, for example.

Though, as long as the coefficients are all integers, they will always have a solution in the form p + q√(n), where p and q are rational and n is an integer. That can be shown by just applying the quadratic formula.

[hazy tendon]

I see

@odd isle Thank you for your assistance. +1 rep ☝️

[glossy night]

@hazy tendon has given 1 rep to @odd isle

[hazy tendon]

@stiff spade Thank you for your assistance. +1 rep 😄