can someone help in math

[Tocsin_786]

so for 3 years that ive been doing algebra ive never gotten one thing in my head and that is how to factor stuff. it never sits 

can somone hlep me out with this problem below. , link me to a good video that explains alll factoring methods and what not.

 

2x^2-2x-24≤0

[Youbetternot]

Ah damn numbers!

[nightlitezzz]

Are you sure it's 2x^2? Btw @Youbetternot, stop spamming useless information and at least try to help.

[Bradles59]

The answer is .... 42

 

I wish I could help but im as clueless as you

[Youbetternot]

if it's 2x^2-2x-24=0  then that I can solve

[T.Vengeance]

It's actually quite easy. Just graph the zeros and when the graph is under the x-axis, it's less than zero.

Since the degree is even and the leading coefficient is positive, the end behaviour as x approaches +/- infinity, y approaches + infinity.

 

To factor a quadratic with a leading coefficient of 1 in the format of ax^2-bx-c, you just have to find 2 numbers that multiply to -c, add up to -b. 

[Tocsin_786]

if it's 2x^2-2x-24=0  then that I can solve

try to solve that i know how to change it back to the greater than and less than stuff

[Tocsin_786]

It's actually quite easy. Just graph the zeros and when the graph is under the x-axis, it's less than zero.

math.png

 

Since the degree is even and the leading coefficient is positive, the end behaviour as x approaches +/- infinity, y approaches + infinity.

it has to be done on a number line. and yea i know the steps. but need help getting the way to factor and all the different methods and ways of factoring down in my head.

[Tocsin_786]

It's actually quite easy. Just graph the zeros and when the graph is under the x-axis, it's less than zero.

math.png

 

Since the degree is even and the leading coefficient is positive, the end behaviour as x approaches +/- infinity, y approaches + infinity.

Thank youuuu. the way i was thinking was how can i get something to multiply to 24 and still add to 2. i had 12 and something in my mind

[piplupgao]

I can but too lazy

[T.Vengeance]

Thank youuuu. the way i was thinking was how can i get something to multiply to 24 and still add to 2. i had 12 and something in my mind

No. it was wrong...LOL...so gimme a sec to edit.

 

K, I just edited it.

[eoso]

Let's see if I still remember...

 

2x^2-2x-24 factors into (2x-8)(x+3).

 

Set to zero, and you have x _<-3 and x_<4 (dunno how to type less than or equal to, and too lazy to find out)

 

You'd want both to be true, so you'd end up with x_<-3. Similar to a boolean && in computer science.

 

Edit: Don't graph, you'll want to know how to do such low level math algebraically.

 

Edit to the Edit: This might be wrong, hang on while I do the problem once more.

 

Editception: Alright, so I remember now. You'll want to solve for zero, then divide the number line into possible portions, then test each portion.

 

Since your zeros are -3 and 4, you have the possible sections of (negative infinity, -3], [-3,4], and [4,infinity).

 

Take some test cases from each possible section: x=-4, x=0,x=5, respectively.

 

Test each case in original equation. 

For x=-4, 2x^2-2x-24 = 16. Since this is not less than or equal to zero, this portion must be incorrect.

For x=0, 2x^2-2x-24 = -24. Since this is less than or equal to zero, this portion is an answer.

For x=5, 2x^2-2x-24 = 16. Since this is not less than or equal to zero, this portion is incorrect.

 

You are left with [-3,4]. Remember that is it possible to have multiple portions that are correct.

[Tocsin_786]

No. it was wrong...LOL...so gimme a sec to edit.

lol. you had me banging my head on my desk.

[T.Vengeance]

lol. you had me banging my head on my desk.

Yeah me failing at math is a common thing. I understand it fully but what I write down is not what I'm thinking. It sucks that I have this problem....

[Youbetternot]

It's actually quite easy. Just graph the zeros and when the graph is under the x-axis, it's less than zero.

math.png

Since the degree is even and the leading coefficient is positive, the end behaviour as x approaches +/- infinity, y approaches + infinity.

 

To factor a quadratic with a leading coefficient of 1 in the format of ax^2-bx-c, you just have to find 2 numbers that multiply to -c, add up to -b. 

Well since he got it cheers @Tocsin_786

[b1khoa]

2x^2-2x-24 <=0

x^2-x-12 <=0

(x-4)(x+3)<=0

x>-4 and x<=4

I think

[Tocsin_786]

than you all. i know its a tech forum but binaries are numbers too lol. anyways im gonna work on this i got a good idea of how to do it. its just that sometimes i cant get the right numbers or the method to do it as im tought how to do several methods.

[eoso]

I just updated my post to include the algebraic way of solving the problem. No problem, don't hesitate to ask for help. I just feel embarrassed that I missed the problem in my first attempt despite having a job tutoring calculus...

[T.Vengeance]

than you all. i know its a tech forum but binaries are numbers too lol. anyways im gonna work on this i got a good idea of how to do it. its just that sometimes i cant get the right numbers or the method to do it as im tought how to do several methods.

If you're trying to factor quadratics with a coefficient greater than 1, you can try the Australian method. I had one teacher that would teach that and it was so amazing and quick compared to what my other teachers taught. 

 

http://teengle.wordpress.com/2007/09/17/australian-factoring/

[Tocsin_786]

I just updated my post to include the algebraic way of solving the problem. No problem, don't hesitate to ask for help. I just feel embarrassed that I missed the problem in my first attempt despite having a job tutoring calculus...

yea helped out tremendously than you sooo much

[Tocsin_786]

If you're trying to factor quadratics with a coefficient greater than 1, you can try the Australian method. I had one teacher that would teach that and it was so amazing and quick compared to what my other teachers taught. 

 

http://teengle.wordpress.com/2007/09/17/australian-factoring/

this is a million times easier. thank you

[T.Vengeance]

this is a million times easier. thank you

No problem. That's exactly what I said when I changed teachers.

[eoso]

yea helped out tremendously than you sooo much

 

Here is the image format, sorry about the messy handwriting.

[iruixos]

I'm surprised no one linked this yet: https://www.wolframalpha.com/input/?i=2x^2-2x-24%E2%89%A40

[Tocsin_786]

 

Here is the image format, sorry about the messy handwriting.

mine is just as bad handwriting lol.