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anonymous

  • 4 years ago

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 :)

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  1. cristiann
    • 4 years ago
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    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

  2. dumbcow
    • 4 years ago
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    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

  3. anonymous
    • 4 years ago
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    oO thank you guys so much ! :)

  4. anonymous
    • 4 years ago
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    Sorry, but how does x^2 + (k-1)x + k^2 - 16 become k^2+3k-10=0 ?

  5. dumbcow
    • 4 years ago
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    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

  6. dumbcow
    • 4 years ago
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    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

  7. anonymous
    • 4 years ago
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    Okay, i get it now :) thanks!

  8. dumbcow
    • 4 years ago
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    :)

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