For what possible integral values of 'k' can this polynomial be factored? x^2+kx+15?

Question

Can't figure this out at all. Could you please show the steps! Thanks....

Answer 1

ok so you know 15 can only be broken down to 5*3, -5*-3, 15*1 and -15*-1
therefore when you break it up
(x+3)(x+5) using FOIL (I assume you know how to do that) k would equal 8. For (x-3)(x-5) k would be -8. Do the same for the set of (x+15)(x+1) and (x-15)(x-1)

Answer 2

If kx^3 + 2x^2 + 2x + 3 and kx^3 - 2x + 9 have a common factor, then their difference also has sme common factor. => 2x^2 + 4x - 6 => x^2 + 2x - 3 has same factor. x^2 + 2x - 3 = (x+3)(x-1). Put x = 1, -3 in any given eqn and equate value = 0 to get values of k. kx^3 + 2x^2 + 2x + 3 = k + 2 + 2 + 3 = 0 => k = -7 for x = 1. and -27k + 18 -6 + 3 = 0 => k = 5/9 for x = -3. kx^3 - 2x + 9 = k - 2 + 9 = 0. => k = -7 for x = 1, -27k + 6 + 9 = 0 => k = 5/9 for k = -3. Thus we get 2 values of k = -7, 5/9.

Answer 3

Any quadratic expression (ax^2+bx+c) can be factorize in real two linear factors only when D>=0.
Here D=b^2-4ac
so x^2+kx+15 can be break in linear factors only when
k^2-60>=0 or you can say k^2>=60
which will give you k<=-sqrt(60) or k>=sqrt(60)
so there will be infinite no of possible integral value of k can be.
more details at http://www.mathiit.in

Answer 4

The answers are +or-8 and +or-16. Those who answered that only positive factors were possible seem to have forgotten that k could be negative.

Answer 5

x^2+kx+15
if
x=-1then
1-k+15=0
k=16
to be factored it should be 16
we get factors as x+1 and x+15
there are many solutions you can get in similar way using x=-3,
-5,-7 and so on

Answer 6

There infinite if you use fractions. If you use whole numbers, only 2 sets.
(x+3)(x+5) and
(x+15)(x+1)

(x+[5/3])(x+9) and
(x+[3/2])(x+10) are some examples of fractions.

Answer 7

In order for it to be factorable, k must equal two numbers 'x' and 'y' such that:
xy = 15
Here are the possibilities
k = 16
k = 8
k = -16
k = -8