# Need help finding the value of a constant

• Dec 27th 2012, 02:03 PM
dfung787
Need help finding the value of a constant
I need help with the foll. problem:

P(x) = x3 + 2x2 - (4K + 5)x - 6 have three distinct zeros. Two of these are identical to the zeros of f(x) = x2+ 5x + K. Compute the value of K.

Attempts:

Using the rational roots theorem:

+-1, +-2, +-3, +-6

I chose -3 because i thought it would work. Then I used synthetic division to find the other zeros.

-3 1 2 (4K + 5) -6
-3 3 6
1 -1 -2 0

(4k+5) must be -5 because -5+3 = -2

next I found the other zeros:

(
x2 - x - 2) = (x + 1) (x - 2)

So, the zeros of this function are: 2, -1, and -3. As stated before, (4K + 5) = - 5

4k + 5 = -5

k = - 10/4 or -2.5

Now, when I substitute -2.5 to
f(x) = x2+ 5x + K, the zeros i get are not the same to the other function...I'm stuck here.

Thanks for helping me ;)

• Dec 27th 2012, 02:16 PM
ILikeSerena
Re: Need help finding the value of a constant
Hi dfung787! :)

You are making this a bit more complicated than it needs to be.
Furthermore, it is not given that the roots are rational numbers, so there is no guarantee that the rational root theorem will help.

The fact that the second order equation has the same roots means that P(x) can be factorized into:
P(x) = (x - A)(x2+ 5x + K)

If you multiply this and equate each coefficient to the coefficients of P(x), what do you get?
• Dec 27th 2012, 03:04 PM
dfung787
Re: Need help finding the value of a constant

When I multiply P(x) = (x - A)(x2+ 5x + K) I get this:

x3 + 5x2 + xK - AX2 - 5Ax - AK

Then I separated each term depending on the exponent:

x3 + 5x2 - AX2 + Kx - 5Ax - AK

I found that A = 3

5 - A = 2
A = 3

and K= 2

AK= -6
-3k = -6
k = 2
• Dec 27th 2012, 03:14 PM
ILikeSerena
Re: Need help finding the value of a constant
Good!

As you can see K=2 is the solution of 2 of the equations.
So the solution is consistent as expected.
Furthermore it confirms you have the proper solution.

Cheers!
• Dec 27th 2012, 03:25 PM
dfung787
Re: Need help finding the value of a constant
Thanks you so much sir!