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 cubic polynomial x^3+2x^2+kx-18.

Question

shubhamsrg · Answer

it means that if f(x) =  x^3+2x^2+kx+3 by x-3
then f(3) = 21 (since x-3 leaves remainder 21 on division)
make that substitution and solve for k.
then you use long division up next to solve for quotient ,,and then the roots.

anonymous · Answer

$$x^3+2x^2+kx+3=(x-3)(x^2+ax+b)+21 \  u \ can \ utilize \ this$$

anonymous · Answer

Oh gosh...this will take a while...
First, set up the synthetic division.
3|   1      2     k             3
              3    15           3(k + 15)
      1      5  (k + 15)     21
Therefore, set up an equation.
3(k + 15) + 3 = 21
3(k + 15) = 18
k + 15 = 6
k = -9

x^3 + 2x^2 - 9x + 3

anonymous · Answer

Now that you have the value of k, use the rational root theorem and Descartes Rule of Signs to decrease the degree to a quadratic.