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MissKelly~
Find all values of k so that each polynomial can be factored using integers. 1.) x^2+kx-19 2.) x^2 - 8x+k, k>0
what means using integers?
Integer roots or integer value of k?
I don't know...I just wrote word for word for what it says in the book...All I know is that for #1 the answers will be in integers...
For the first one: -18, 18.
How do you solve it? Distributive...factor?
Do you know Vieta's Theorem?
Not yet, The questions I got was from the book on the chapter of Quadratic equations, and it said it required Higher-order thinking skills...
So you factor it out into 2 binomials?
Vieta's Theorem is not so hard to get. I advice you to get it and try to use for your task.
I'm not so sure my teacher wants me to apply Vieta's Theorem yet, because she hates it when we move on ahead of her teachings...btw she can't teach =_=
Well I have to leave for school in a little while, there's a chemical equation quiz waiting for me...
Thanks for helping me~
Hehe why is that? Well I'm sorry I have to go to school now..
See you guys around^^ Bye
If you want to factor \(x^2+kx-19\), I hope you will think it will be \((x-a)(x-b)\). So, when expanding: \(x^2-(a+b)x+ab\). Now just look at the coefficients. \(k=-(a+b), -19=ab\). You want only integers. From the second equality it will be only \(a=1,b=-19\) or \(a=19,b=-1\), because 19 is a prime number. Now try to find \(k\) in both cases.
Just try to explain. Not so sure he got it.
If \(-(a+b)=-8\), so \(a+b=8\). And \(ab=k\), we get \(a(8-a)=k>0\). It will be if \(-a^2+8a>0\).|dw:1357824979524:dw| Answer: 1, 2, 3, 4, 5, 6, 7.
Oops. Sorry, it's not right.
The answer will be if you put all this numbers: 1, 2, 3, 4, 5, 6, 7 in the expression for \(k=a(8-a)\) instead of \(a\).
How did you get that this will be all the possible values for \(k\), that you've written k = {0, 7, 12, 15, 16}? \(k=0\) doesn't suit.
\(k>0\) in the statement.
I can't be absolutely sure that this will be ALL the values for \(k\) by just guessing its value.
The source of our misunderstanding is my bad English. Sometimes I can't get what is spoken about.
You say that it is no need in writing \(k=a(8-a), a=1,2,\ldots ,7\). Because it will have the same value for \(a=1\) and \(a=7\). I just showed the way I had solved this task. Sorry.
Okay I'm back...whoooooa. And yeah I got it now^^