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anonymous
 4 years ago
Help please, How do you find the value of k of which x^2 + (k1)x + k^2  16 is exactly divisible by (x3) but not divisible by (x+4) ? thank you :)
anonymous
 4 years ago
Help please, How do you find the value of k of which x^2 + (k1)x + k^2  16 is exactly divisible by (x3) but not divisible by (x+4) ? thank you :)

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cristiann
 4 years ago
Best ResponseYou've already chosen the best response.2For the polynomial p(x,k)=x^2 + (k1)x + k^2  16 to be divisible by x3 it is necessary that p(3)=0, which leads to k^2+3k10=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
 4 years ago
Best ResponseYou've already chosen the best response.0Let c be some number. Then you want the quadratic to be factored with x3 as a factor but not x+4 (x3)(xc) = x^2 +(k1)x +k^216 ...hmm your way is easier ^^ but yes k=5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oO thank you guys so much ! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, but how does x^2 + (k1)x + k^2  16 become k^2+3k10=0 ?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0Expand left side x^2 (c+3)x +3c = x^2 +(k1)x +k^216 > (c+3) = k1 solve for c c = k2 > 3c = k^2 16 substitute in for c 3k6 = k^2 16 > k^2 +3k 10 = 0

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0oh oops how cristiann came up with it is by plugging in 3 for x p(3) = 0 3^2 +3(k1) +k^216 = 0 9+3k3 +k^2 16 = 0 k^2 +3k 10 = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, i get it now :) thanks!
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