# Integral values

• Oct 19th 2012, 08:06 PM
ford2008
Integral values
Hey guys,

the task: For what integral values of k can 4x^2 + kx - 3 be factored?

(4) (-3) = -12 Factors of 12= 1,12 2,6 and 3,4

So are the possible values for k = +11, +4 and +1 or all negative?

They should be positive right, because I am looking for k and the sign is positive in front of k.
• Oct 19th 2012, 08:13 PM
MarkFL
Re: Integral values
You may either have:

$(2x+1)(2x-3)$ or $(2x-1)(2x+3)$

or

$(4x+1)(x-3)$ or $(4x-1)(x+3)$
• Oct 19th 2012, 09:19 PM
ford2008
Re: Integral values
I think it should be

(2x-1) (2x+3) or (2x+3)(2x-1)

or

(4x-1)(x+3) or (4x-3)(x+1)

because

(2x+1)(2x-3) = 4x^2 -4x-3 (in that case the value k would be negative) It should be 4x^2 + kx -3 (positive).
• Oct 19th 2012, 09:48 PM
MarkFL
Re: Integral values
I missed a pair:

$(4x+3)(x-1)$ or $(4x-3)(x+1)$

$k$ is just an arbitrary integer, it may either be positive or negative.
• Oct 20th 2012, 10:18 AM
ford2008
Re: Integral values
Okay, I got that.

So, for what integral values of k can 4x^2 + kx + 3 be factored? (note +3, not -3 like in task 1).
I would like to know your answer, because I got an example for 4x^2 + kx + 3 in my book. There is nothing about that k is just an arbitrary integer and it may be either positive or negative.
• Oct 20th 2012, 11:44 AM
MarkFL
Re: Integral values
When it is said "for what integral values of $k$" this implies that $k$ can be negative or positive, since the integers are {...-2,-1,0,1,2,...}.

The possibilities here are:

$(4x+1)(x+3)$ so $k=13$

$(4x-1)(x-3)$ so $k=-13$

$(4x+3)(x+1)$ so $k=7$

$(4x-3)(x-1)$ so $k=-7$

$(2x+1)(2x+3)$ so $k=8$

$(2x-1)(2x-3)$ so $k=-8$
• Oct 21st 2012, 02:23 AM
Farenxdesign
Re: Integral values
I hate thiss..

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• Oct 21st 2012, 09:38 AM
Soroban
Re: Integral values
Hello, ford2008!

Quote:

For what integral values of $k$ can $4x^2 + kx - 3$ be factored?

(4) (-3) = -12 . . Factors of 12: (1,12), (2,6) and (3,4).

So are the possible values are: . $k \;=\;\pm1,\,\pm4,\,\pm11$
This is correct!

My approach . . .

We have: . $4x^2 + kx - 3 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{-k \pm\sqrt{k^2 + 48}}{8}$

To be factorable, the discriminant must be a square:
. . $\text{that is: }\:k^2+48 \:=\:a^2\,\text{ for some integer }a.$

This occurs when: . $k \;=\;\pm1,\,\pm4,\,\pm11$