anonymous
  • anonymous
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 :)
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

cristiann
  • cristiann
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
dumbcow
  • dumbcow
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
anonymous
  • anonymous
oO thank you guys so much ! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Sorry, but how does x^2 + (k-1)x + k^2 - 16 become k^2+3k-10=0 ?
dumbcow
  • dumbcow
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
dumbcow
  • dumbcow
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
anonymous
  • anonymous
Okay, i get it now :) thanks!
dumbcow
  • dumbcow
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.