SOLUTION: If {{{x^2+kx+k+1=(x+p)(x+2)}}} for all values of x and k and p are constants, what is the value of k? A)5 B)4 c)3 d)2

Algebra ->  Functions -> SOLUTION: If {{{x^2+kx+k+1=(x+p)(x+2)}}} for all values of x and k and p are constants, what is the value of k? A)5 B)4 c)3 d)2       Log On


   

Question 1041226: If x%5E2%2Bkx%2Bk%2B1=%28x%2Bp%29%28x%2B2%29 for all values of x and k and p are constants, what is the value of k?
A)5
B)4
c)3
d)2
Found 2 solutions by rothauserc, Edwin McCravy:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
x^2 +kx +(k+1) = x^2 +2x +px +2p = x^2 +(2+p)x +2p
:
we have two equations in two unknowns
:
1) k+1 = 2p
2) k = 2+p
:
substitute for k in equation 1
:
2+p+1 = 2p
:
p = 3
:
************
k = 2+3 = 5
answer is A)
************
:

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2%2Bkx%2Bk%2B1%22%22=%22%22%28x%2Bp%29%28x%2B2%29

Let x = 0

0%5E2%2Bk%2A0%2Bk%2B1%22%22=%22%22%280%2Bp%29%280%2B2%29

k%2B1%22%22=%22%22%28p%29%282%29

k%2B1%22%22=%22%222p

Let x = 1

1%5E2%2Bk%2A1%2Bk%2B1%22%22=%22%22%281%2Bp%29%281%2B2%29

1%2Bk%2Bk%2B1%22%22=%22%22%281%2Bp%29%281%2B2%29

2%2B2k%22%22=%22%221%2B2%2B2p%2Bp

2%2B2k%22%22=%22%223p%2B3

2k%22%22=%22%223p%2B1

So we solve the system of equations:

system%28k%2B1=2p%2C2k+=+3p%2B1%29 

Solve by substitution and get k=5, p=3.

Edwin