Question

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

Answer 1

what means using integers?

Answer 2

Integer roots or integer value of k?

Answer 3

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...

Answer 4

For the first one: -18, 18.

Answer 5

How do you solve it? Distributive...factor?

Answer 6

Do you know Vieta's Theorem?

Answer 7

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...

Answer 8

So you factor it out into 2 binomials?

Answer 9

Vieta's Theorem is not so hard to get. I advice you to get it and try to use for your task.

Answer 10

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 =_=

Answer 11

Well I have to leave for school in a little while, there's a chemical equation quiz waiting for me...

Answer 12

Thanks for helping me~

Answer 13

Hehe why is that? Well I'm sorry I have to go to school now..

Answer 14

See you guys around^^ Bye

Answer 15

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.

Answer 16

Just try to explain. Not so sure he got it.

Answer 17

Or she.

Answer 18

If \(-(a+b)=-8\), so \(a+b=8\). And \(ab=k\), we get \(a(8-a)=k>0\). It will be if \(-a^2+8a>0\).
Answer: 1, 2, 3, 4, 5, 6, 7.