Question

If the remainder on division of x3 + 2x2 + kx +3 by x – 3 is 21, find the quotient and the
value of k. Hence, find the zeroes of the cubic polynomial x3 + 2x2 + kx – 18.

Answer 1

Answer:

Step-by-step explanation:

  Given x^3 + 2 x^2 + k x + 3

  X – 3 = 0

  X = 3

  f(3) = 21

  3^3 + 2(3)^2 + k(3) + 3 = 21

 27 + 18 + 3k + 3 = 21

 3k = - 27

 K = - 9

Now  

 X^3 + 2 x^2 + k x – 18

 We have k = -9

 X^3 + 2 x^2 – 9 x + 3

So by division we get  

 X – 3)x + 2 x^2 – 9 x + 3(x^2 + 5 x + 6

          X^3 – 3 x^2

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                   5 x^2 – 9 x + 3

                    5 x^2 – 15 x

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                                  6 x + 3

                                  6 x – 18

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                                       21

So we get x^3 + 2 x^2 – 9 x + 3 = (x – 3)(x^2 + 5 x + 6) + 21

 (x – 3)(x^2 + 5 x + 6)

 (x – 3)(x^2 + 2 x + 3 x + 6)

 (x – 3)(x(x + 2) + 3(x + 2)

 (x – 3)(x + 2)(x + 3)

So zeroes of x^3 + 2 x^2 – 9 x – 18 are 3, - 2 and – 3