What is the factorization of 4x^2-31x+21?
please don't leave, we want to help you. Many people like (4x-3)(x-7)
people like you left OS, because of small (as they though) inconveniences.
@MaxRide
I'm not planning on leaving if this helps me.
Good, b/c it does help you :) See, how I am tagging you ? You welcome!
Um, aren't the -3 and -7 supposed to add up to -31?
( 4 x - 3 ) (x - 7 ) = 4 x ( x - 7 ) - 3 ( x - 7 ) = 4x^2-28x - 3x + 21 = 4x^2-31x + 21 .
-3 and -7 when you multiply , the middle terms are supposed to add up to 31. also, -3 and -7 are suppose to be factors of 21, and yes, (-3) * (-7)= 21
Hang on one second.
Sure....................
Ok, now I feel completely dumb. I forgot about the 4.
it's okay... so do you get it, or need a little more help ?
I think I get it, I'll try a few more problems and be back if I need more help.
Thank you so much!
You welcome, it is totally my pleasure, and my practice :)
Haha.
☻
I'm doing problems 34 to 51, so instead of typing a problem out, I'll say the number, it you don't mind.
If I'm wrong, I'll sound stupid, but is 34 prime?
Never mind.
34 is not prime, (3x+1)(2x+7) is the factorization... also, if the discriminant is a perfect square than it is factor-able and if the discriminant is not a perfect square then it is not factor-able.
Which number is the discriminant?
have you ever worked with a quadratic formula ?
I don't believe so.
I will show you it fast... \(\Large\color{blue}{ \bf -3x-x+2=0 }\) is the quadratic equation. One of the approaches to solve is it, is the quadratic formula, below in red \(\Large\color{red}{ \bf x= \frac{ \sqrt{b^2-4ac} }{2a} }\)
the discriminant is the number inside the square root
what are the abc ? ax^2 +bx+c=0 is the standard form, COMPARE... -3x -x +2=0 so a=-3 b=-1 c=2
So your problem.. in order to say that it is unfactorable, you must show that b^2-4ac (in your case, (-1)^2 - 4(-3)(2) ) is not a perfect square, in other words, that it won't equal an integer after the root is taken.
(-1)^2 - 4(-3)(2) = 1 + 24 = 25. we know that √25= ±5, so it's therefore a perfect square and therefore -3x^2-x+2 is NOT PRIME.
Where did you get the (-1)^2 - 4(-3)(2) from?
Was that from my problem?
-1 is "b" (-3) is "a" 2 is "c" yes, we are doing 51, aren't we ?
Sorry, I was working on some problem with my dad while you were explaining that and lost track of which problem we were on. My bad. Oops!
a, b, and c, is referring to the red formula I posted before... the part inside the square root. Which is, b²-4ac
Ok, I see it now.
So, your polynomial is factorable, in 51, right ?
trinomial
Yes.
Kay, I'm guessing I'm wrong.
OK, so lets think how to factor it.... - 3x² - x + 2 = 0 the approach I am going to take is to factor it out of -1 first to make it look better ;) 3x² + x - 2 = 0 (-1)(3x - 2 ) (x +1 )=0 so knowing that, we can multiply either of the factors ( "(3x - 2 )" or "(x +1 )" times -1 ) so the answer can be (- 3x + 2 ) (x +1 ) or ( 3x - 2 ) ( -x - 1 )
Are you just adding the =0 to the end for my/your sake, or is that supposed to go on the end?
for our sake :) to have a better conceptual understanding of how I can factor out of -1.
The answers I already said, and there don't involve any zeros what so ever.
I'm gunna type it out, tell me if I'm wrong, please.
What are you going to type out ?
-3x^2-x+1 -1(3x^2+x-2) Ignore the -1 for now. Ok, I'm officially lost, where do I go from here?
then I am just factoring it....
I am factoring "(3x^2+x-2)" = (3x - 2 ) (x +1 ). Now, I am saying that you can multiply either one of the factors ( "(3x - 2 )" or "(x +1)" ) by -1, (to get the -1 back) and get the final answer. (the rule to understand this would be that -1(a)(b)=(-a)(b) or (a)(-b) ) so, it would therefore be (-a)(b) or (a)(-b) in your case, (- 3x + 2 ) (x +1 ) or ( 3x - 2 ) ( -x - 1 )
I think I've got it. If I'm wrong your going to think I'm the most stupid person on the planet. My final answer would be -1(3x-2)(x+1), right?
well, it is not wrong, it is just that it is technically never written like this.
Ok. As long as it's not wrong, I'm good for now.
If I'm annoying you with my questions, let me know. It's just sometimes hard for my parents to understand what is frustrating me. Thanks for sticking around for so long.
Um, if I finish before you leave, would you be willing to check the even numbers for me? The odds are in the book. If not, that's ok, I'll check them on Monday with my teacher.
@student_basil
you aren't annoying me, I am just making a small tutorial at the moment:)
Ok.
I am done.
Cool beans.
Could you take a peek at #40? Thnx.
Ok... tell me if the discriminant of this is a perfect square or not. In other words can the trinomial in num 40 be factored or not.
No?
11²-(4)(6)(4) 121-96 25 is 25 a perfect square ?
Yes.
Where did the 11^2 come from?
remember the "discriminant in the quadratic formula, b²-4ac ?
Right, got it.
yes, so you have a couple ways to factor it. way 1: (if you really not in the mood to think) set the trinomial equal to zero and solve for x, lets call the number you get for x, "h" and "k" so the factored trinomial can be written as (x-h)(x-k)
you can solve for x (in way 1, ) using a quadratic formula, \(\Huge\color{blue}{ \bf x=~\frac{-b±\sqrt{b^{2}-4ac} }{2a} }\)
tell me if you can do way 1 or not...
Give me a minute to attempt it.
1 question first, what's the little sign after the -b? On second thought, maybe that's not the best one for me to try.
it is ± (plus or minus) when you say ±4 you mean "positive or negative 4" in the formula, it is there to say that the root whatever you get, let say 16, can be either +16 or -16.
Ok. One moment.
So I have -11+/- 5 over 12. Where do I go from there?
yes, so you have -11 +- 5 -------- 12
so either -11 - 5 -------- 12 which gives -4/3 or -11+ 5 -------- 12 which gives 1/2
6x²+11x+4 (2x+1)(3x+4)
-11+5 -6 ------- would be ------ 12 12 thats not 1/2, is it?
yeah, you caught the err... - 1/2 Now look at the numbers in (one half ) (four thirds) but it's in a different order in each parenthesis.
(2x+1)(3x+4)
I gtg now -:( I'll be back :)
alright.
http://openstudy.com/users/student_basil#/updates/53542b8be4b066a1cac3b1a0 check it out, and see if you need any of the symbols there... BYE !
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