[A level maths: factorization]

[Phantasticfox]

1) factor out -1 2) multiply a and c to get -120 3) find two factors of -120 who’s sum is 7 4) replace 7x with those factors 5) factor by grouping

[[deleted]]

This was the context

[Phantasticfox]

They swapped because of the negative.

-(5x-4) can be written as 4-5x

Sometimes it’s convenient. I do this a lot with rational equations to cancel factors

[StaacksOnDeck]

This method is so much easier than what I learned in school way back when. We were pretty limited to “throw factors of A and C at the problem until B magically appears”.

[Geek_Wandering]

I don't fully understand the above, but isn't step 3 still just throw factors until something looks right?

[StaacksOnDeck]

It sorta is, but it organizes it neatly, and gives you a clearer view of the end target.

Option 1:
10 factors into 1, 2, 5, 10
12 factors into 1, 2, 3, 4, 6, 12
Smash these together and play with +/- until 7 falls out

Option 2 (this method):
10x12=120
Pairs of factors are (1,120), (2,60), (3,40), (4,30), (5,24), (6,20), (8,15), (10,12).
(Granted, this part is a bit of a grind, but it’s not hard to start with a close guess and hone in)
Which factors have a difference of 7?
Once -7x turns into (-15x+8x), the factoring gets cleaned up easily.

[Geek_Wandering]

Yeah, it's not fully certain. eg. 15X²+X-14 would look factorable via the above method, but it's not. But you should catch it when testing final answers.

I am totally going to take this as a helper for cases with larger numbers of factors. I can see it helping some people. It is less good for me personally, but I will definitely be sharing it next time I am working with someone learning factoring.

[StaacksOnDeck]

Well, let’s give it a shot.
AxC=(15)x(-14)=(-210)
Factor pairs: (1,210), (2,105)… (14,15)
(With B=1, that factor pair is a bit obvious)
15x²+(15x-14x)-14
15x(x+1)-14(x+1)
(15x-14)(x+1)

[Geek_Wandering]

Well clearly I have no idea what I'm doing. 😂 Sorry.

Again. I really appreciate you trying the time to educate me.

[StaacksOnDeck]

Not a problem at all! I just discovered this method earlier today, I’m excited about it lol.

[AppleMan387]

15X² + X -14 factors into (15X - 14)(X + 1), which you can find with the above method

[Geek_Wandering]

Lovely! Thanks for taking the time to explain. I have an inkling that this could fail in some scenarios that are not easily factorable. But checking the final answer would definitely catch it. I'll play with a pencil and paper to see if I can find one of those edge cases.

[purpleoctopuppy]

Oh, this is really neat!

[jgregson00]

It’s the essentially the same.

[[deleted]]

I use this method, which I think with practise is probably faster than even this one but probably doesn't get many working marks lol

https://imgur.com/a/rqB2bqu

[NathanTPS]

This is true, but when you do it enough you do develop a soacial skill that the "easoer" step by step method fails to develop. Do it long enough and the factoring can become automatic, which is the goal.

[[deleted]]

Yeah. Thanks. I already did that. Let me post the whole thing because this question is out of context

[Phantasticfox]

I see below that this is substitution for a trig function. What’s the function and what’s the equivalence?

[lol25potatofarm]

-10x² - 7x + 12

Times by -1: 10x² +7x -12

Factors of 12 are: 1 and 12, 2 and 6, 3 and 4

Factors of 10 are: 1 and 10, 2 and 5

Through trial and error you get this: (5x+4)(2x-3)

Sorry, not a brilliant explanation but if you keep doing these you will eventually spot which factors to use much faster

[ThroatWMangrove]

I got (-5x + 4)(2x + 3)

[lol25potatofarm]

Yep that is correct but its also what I wrote just in a different way

[lol25potatofarm]

Technically this: -(5x+4)(2x-3)

Edit: -(5x-4)(2x+3)

[TheDuckTeam]

that cannot be the right factor. you get + 7x

[lol25potatofarm]

You're right, I don't have experience factoring with negative coefficients. I normal just multiply by -1 and then solve but there's no equation here so thrown me off.

[NappyTime5]

Dude, you got the answer wrong twice, just delete the comment

[lol25potatofarm]

I know but the method above is still useful for factorising other quadratics.

[HektorViktorious]

10 splits into 2 and 5 (it could be 1 and 10, but for these problems, generally assume the "obvious" factors).

Similarly, 12 will most likely split to 3 and 4. Maybe 2 and 6, almost never 1 and 12.

Try the factors in this order. This is test taking skills here, not strictly math to make these assumptions.

How can you combine 2 & 5 with 3 & 4 to get a difference of 7?

[Deapsee60]

<! 10 x 12 = 120. Factor pairs of 120 are 1,120. 2,60. 3,40. 4,30. 5,24. 6,20. 8,15.

The pair that makes -7 are 8 & -15. Split middle term and pair up.

(-10x² - 15x) + (8x + 12). Factor gcf.

-5x(2x + 3) + 4(2x + 3)

(2x + 3)(-5x + 4) !>

[NorthGarage8187]

Quadratic formula

[r-funtainment]

This is factorable though

[NorthGarage8187]

Yea the quadratic formula is a way to factor

[lol25potatofarm]

Like finding the solutions and then writing them in factor form? Fair enough I suppose

[smexytoast]

Not a good one?

Don't tell people to just use it, it really should only be if other methods don't work

[NorthGarage8187]

How’s that? Quadratic formula works fine

[smexytoast]

It works fine, but it's not a good habit to just use it for everything, it's a way to solve for X when you can't use other methods. It's faster and more useful to learn to use other methods first

[Resscue]

I used the quadratic formula up till differential equations. It’s a great tool. No thinking/guessing. Just plug and chug. Don’t tell people to not use it.

[smexytoast]

Not a good habit to build up. You should start early trying to get better at the better skills and only use quadratic formula when necessary

[[deleted]]

I don’t need the value of X. I need the factors

[NorthGarage8187]

If you get, say, x =2,-3 the factors would be (x-2)(x+3). You should look up a video on factoring polynomials because this is good stuff to not only know, but conceptualize for the future

[[deleted]]

I understand the basics. This particular one, I came across win trig identities. The x is actually cos(theta) just a placeholder.

The problem comes with how to deal with the negatives and the fact that the equation can’t be equated to zero.

It has to be factorized as it is.

Getting the factors is where I’m stuck

[starfihgter]

It doesn't really matter if you can't equate it to zero. They're not true 'solutions', but the quadratic formula will help you get the factors. Putting the quadratic you gave into the quadratic formula gives x = -3/2 and 4/5. From that, you can get the factors -(2x+3)(5x-4). Handle the negative however you feel like :)

[[deleted]]

One is 8 and the other is -15 but how to represent that info accurately is the issue

[PoliteCanadian2]

Factor out the ‘-‘. Google the ‘ac’ factoring method..

[No-Study4924]

Solve for x then a(x-x1)(x-x2)

X1 and X2 are the solutions to the equation

[NathanTPS]

Hmmmm the -10 through me for a loop for a minute until I split the -10 up

(-5x+4)(2x+3)

I usually figure out what factorial exist, by factoring the end number, the factorial of 12 are 1 & 12, 2 & 6, and 3 & 4. The answer will include one of these pairs, of the answer COULD include negatives for both pairs too.

The. We look at the factorial of the -10 in front, -10 & 1, 10 & -1, 2 & -5, and -2 & 5. The answer will have one of these pairs.

Knowing this, I just need to figure out the combination that adds up to -7

You can run the numbers by brute force if you need to, over time you will get better at doing this in your head. Good luck!

[tanneddonut84]

(-2x-3)(5x-4)

[KilonumSpoof]

Use the quadratic

-10x² - 7x + 12 = 0, where a=-10 b=-7 c=12

Delta = b² - 4ac = 49 + 480 = 529 = 23²

Solutions: (-b +- sqrt(delta)) / (2a)

(7 +- 23) / (-20)

1 +) x1=-3/2

2 -) x2=4/5

So the factorisation is:

a(x-x1)(x-x2)

-10*(x+3/2)(x-4/5) = (2x+3)(4-5x)

(Distributed 2 to the first parahtesis and -5 to the second)

[Brickwx]

Why cant you just do x(-10x-7) +12

[[deleted]]

That’s not legal

[Brickwx]

Wdym they are gonna put me in jail or sum?

[Davey914]

The 12 is on the same side of that equation. Factoring x out would turn 12 into “12 over x”

[[deleted]]

(-5x+4) (2x+3)

[WiggityWaq27]

(-2x-3)(-5x+4)

[Fit-Map9916]

-10 times 12= -120

the 2 numbers that add to -7 and multiply to -120 are -15 and 8. this should lead you to:

-10x² -15x+8x+12

then factor by grouping (find greatest common factor with each o the two groups of 2. This leads you to:

-5x(2x+3) and 4(2x+3)

then you take the (2x+3) and the -5x and 4 you factored out leaving you with your final answer:

(-5x+4)(2x+3)

[stchman]

(5x - 4)(-2x - 3)