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galben [10]
3 years ago
7

if the remainder on division of x^3+2x^2+kx+3 by x-3 is 21,find the quotient and the value of k. Hence,find the zeroes of the cu

bic polynomial x^3+2x^2+kx-18.
Mathematics
1 answer:
maxonik [38]3 years ago
8 0
Let p(x)=x^3+2x^2+kx+3 
On dividing p(x) by x-3, the remainder is 21. Therefore,
 P(3)=21
 Substituting x=3 in p(x)
 P(3)=3^3 +2*(3)^2+k*3+3
  =27+18+3k+3
  =48+3k
 We know that, p(3)=21.
 So, 48+3k=21
  3k=21-48
  3k=-27
  k=-27/3 =-9
 now, p(x) =x^3+2x^2-9x-18 
 -2 is a factor of p(x) on inspection. Therefore, divide p(x) by x+2 to find the
zeroes of the polynomial.
 On dividing, we get the factors to be, (x^2-9)(x+2)
 (x^2-3^2)(x+2)
 Factorizing using the identity a^2-b^2=(a+b)(a-b) we get,
 (x+3)(x-3)(x+2)
 Therefore, the zeroes of the polynomials are -3,+3 and -2.
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b) Standard Deviation:

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3 years ago
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Wewaii [24]
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3 years ago
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uysha [10]

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CD = DA - CA

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