Question

-2x^2-4x+6
I get -2(x-1)(x+3)
Can someone tell me where my mistake is?

Answer 1

that's correct, why do you think it's wrong?

Answer 2

if (-1) is multiplied to (x-1), you'll get
\[2(1-x)(x+3)\]

Answer 3

that's true sirm3d, but what goldcoastgeekgirl got is a perfectly valid answer

Answer 4

so either there's a typo somewhere or there's more to this problem

Answer 5

It's a review question in my text book. Oops. That is the correct answer. I get (x-2)
:)

Answer 6

oh so you got (x-2) in there by mistake?

Answer 7

The correct answer is -2(x-1)(x+3)

Answer 8

yes

Answer 9

but I get -2(x-2)(x+3)

Answer 10

ah i gotcha

Answer 11

do you see how to get the answer now?

Answer 12

or are you still unsure how they got that

Answer 13

still confused :(

Answer 14

ok first you factor out -2 to get this

-2x^2-4x+6

-2(x^2+2x-3)

Answer 15

now focus on factoring the stuff inside

so focus on factoring x^2+2x-3

Answer 16

to do this, you find two numbers that

a) multiply to -3 (last term)

AND

b) add to the middle coefficient 2

Answer 17

ok

Answer 18

what do you get?

Answer 19

Yes, 3 and 1

Answer 20

close, but not quite

Answer 21

My confusion I think comes from learning (trying to) too much too quickly.
I was trying to find something that factored into six but when put in (x ) (x ) equaled -4

Answer 22

But maybe you do that with a different sort of equation?

Answer 23

well once you pull out the -2, the -6 drops to a -3

Answer 24

I meant the +6 becomes -3

Answer 25

notice how if you distribute back, you'll get -6 back again

Answer 26

and does (ax-2ax+2a^2) apply here?

Answer 27

errr a^2 not ax

Answer 28

that's the perfect square form

Answer 29

in this case, the answer is no

Answer 30

sigh...... I'll just go bang my head against a wall for bit.
I'll never figure this out.

Answer 31

yes you will, just need to practice some more

Answer 32

as my profile says, I

Answer 33

I'm old, but I love maths and am determined to master algebra, but my daughter says it's time I accepted the fact that 'Maths is just not that into you, ma."

Answer 34

well you never know, no matter who tells you that

Answer 35

Why not factor out -2x( why just -2( ?

Answer 36

because the +6 at the end doesn't have an x on it

Answer 37

Ok. Some of the equations I've worked on lately only concern themselves with the first two parts of the equation, but obviously that is for different goals.

Answer 38

what do you mean

Answer 39

brb, I'll grab an example

Answer 40

alright

Answer 41

can't find one but earlier this week, I had to work out something and the process was to put (x ) (x ) under the equation, choose two factors from the third part of the equation, and through elimination work out if it was, for example ( x+2) (x-3) or( x+3) (x-2) and this was confirmed if the inner part of the two sets of backets +2) (x when added or subtracted from the out part (x )( -3) were equal to the second term of the original equation.
sheeesh, that is probably all double dutch... it certainly is an algebra soup in my head.

Answer 42

i see, well the method we're using is sorta like the method you are describing

Answer 43

basically we have x^2+2x-3

and we want it in the form

(x + ??)(x + ??)

Answer 44

the numbers that go in the spaces of ?? are the numbers that multiply to -3 and add to 2

Answer 45

ok, and how do you know, by looking that the answer will be in that format?

Answer 46

because the leading coefficient of x^2+2x-3 is 1

Answer 47

if it was something like 2x^2+2x-3, then it wouldn't be in that form

Answer 48

ok, well I think here is the crux of my confusion.

Answer 49

I

Answer 50

look at the question and go ... where do I start?

Answer 51

you should always try to factor out the GCF

Answer 52

that should be your first step

Answer 53

ok.. always the first port of call.
That's a good pointer from me. Thanks

Answer 54

yw

Answer 55

from there, if the stuff on the inside has a leading coefficient of 1, then you use this method

Answer 56

Ok, now this is probably going to sound confusing too.
Some problems have, for example the coefficients 16 8 4 (just pulled numbers out of the air).
So lets say the common factors are 2 and 4. Sometimes your answer will be 12 4 1 because you subract the factor, other times it's 4 2 1.
I think I missed some very fundamental but small piece of information when I was first learning this and never found it.

Answer 57

well for something like that, maybe something like this will help

http://www.regentsprep.org/Regents/math/algtrig/ATV1/Ltri3.htm

it's a tutorial on the AC method (another way of factoring, but it allows leading coefficients of any number, and not just 1)

Answer 58

Thank you jim. it's really refreshing to talk to a person. I've watched my lectures, and hundreds of googled vids, but it's not the same as having someone say 'that is where you went wrong' etc.
Much appreciated.

Answer 59

you're welcome

Answer 60

Wow, that AC method looks interesting.

Answer 61

yes it's very handy

Answer 62

I will bookmark this site. I think I will be a regular visitor. Do you blacklist people if they are too thick? :)

Answer 63

no you're not thick at all, so no worries

Answer 64

Thanks Jim. where in the world are you?

Answer 65

the US

Answer 66

I wish I'd found this site a few days ago. I'm committed to paying a tutor $50 tomorrow for an hour with a tutor.

Answer 67

well an in person tutor may be better, who knows

Answer 68

esp if she or he is good

Answer 69

so out of curiosity, if the equation had been , um 12x + ax - z what form would the answer have taken?

Answer 70

the form would chance to

(??x + ??)(??x + ??)

something like that

Answer 71

using variables it would look like this

(ax+b)(cx+d)

Answer 72

change* not chance...can't spell

Answer 73

ok, and Factor 16-x^2 (4-x) (4+x) I get the first half, but how do they come to 4+x for the second?

Answer 74

you use the special rule

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

Answer 75

in this case,

a = 4
b = x

Answer 76

ok. it's a case of 'this is how it's done. do it'
But what if it's a negative? I looked for a rule but couldn't find one.

Answer 77

what do you mean

Answer 78

laptop froze. grrrr

ok, on my cheatsheet it says (ab)^n = a^n b^n but doesn't tell you what to do with -(ab) or (-ab)
And if there is no 'rule' for it I'm lost

Answer 79

so you mean -(ab)^n ?

Answer 80

yes, but that's just an example. Very few of the 'rules' explain what to do when the problem has negatives.

Answer 81

oh, well you try to factor them out if they bug you

Answer 82

so something

-2x^2-4x+6

turns into

-2(x^2 + 2x - 3)

because we factor out -2

Answer 83

and we just focus on x^2 + 2x - 3

Answer 84

what I mean is, there is a rule for (ab)^n is there a rule for -(ab)^n. Or is it just presumed that you treat the negative as you would in any equation.
And does that go for most rules?

Answer 85

factor it out? Where did it go? Doesn't that change the whole equation?

Answer 86

when you factor it out, you temporary ignore it while you focus on x^2 + 2x - 3

Answer 87

so it doesn't actually have to have any common factor with any other part of the equation... you can just put it away so long as you remember to bring it back at the end?

Answer 88

exactly

Answer 89

ah, a lightbulb moment. :D

Answer 90

lol that's great

Answer 91

well, I have to go watch the second lecture on absolute values before bed. That is really screwing with my already shaky grasp of how to deal with negative lol.

Answer 92

Thanks for your help Jim

Answer 93

you're welcome

Answer 94

All those questions I'm seeing on the left, are they mostly people needing help, or people posing questions for others to solve?

Answer 95

it could be both, but most of the time it's people asking for help

Answer 96

and are you a teacher or just a math enthusiast?

Answer 97

a student learning to become a teacher

Answer 98

so I guess a bit of both