Question

Help please, How do you find the value of k of which x^2 + (k-1)x + k^2 - 16 is exactly divisible by (x-3) but not divisible by (x+4) ? thank you :)

Answer 1

For the polynomial
p(x,k)=x^2 + (k-1)x + k^2 - 16
to be divisible by x-3 it is necessary that p(3)=0, which leads to k^2+3k-10=0
with 2 solutions, k=2 and k=-5
p(-4,2)=0 and p(-4,-5) not =0 so the solution is k=-5

Answer 2

Let c be some number.
Then you want the quadratic to be factored with x-3 as a factor but not x+4

(x-3)(x-c) = x^2 +(k-1)x +k^2-16

...hmm your way is easier ^^

but yes k=-5

Answer 3

oO thank you guys so much ! :)

Answer 4

Sorry, but how does
x^2 + (k-1)x + k^2 - 16
become
k^2+3k-10=0 ?

Answer 5

Expand left side
x^2 -(c+3)x +3c = x^2 +(k-1)x +k^2-16

--> -(c+3) = k-1
solve for c
c = -k-2

--> 3c = k^2 -16
substitute in for c
-3k-6 = k^2 -16

--> k^2 +3k -10 = 0

Answer 6

oh oops
how cristiann came up with it is by plugging in 3 for x
p(3) = 0

3^2 +3(k-1) +k^2-16 = 0
9+3k-3 +k^2 -16 = 0
k^2 +3k -10 = 0

Answer 7

Okay, i get it now :) thanks!

Answer 8

:)