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 zeros of the cubic polynomial x3+2x2+kx-18

Answer 1

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Answer 2

$let \: p(x) = {x}^{3} + 2 {x}^{2} + kx + 3 \\ \: \: \: \: \: \: r(x) = 21 \\ \: \: \: \: \: \: g(x) = x - 3 \\ \\ we \: know \: that \\ \\ p(x) = g(x).q(x) + r(x) \\ \\on \: putting \: there \: values \\ \\ {x}^{3} + 2 {x}^{2} + kx + 3 = (x - 3).q(x) + 21 \\ (x - 3).q(x) = {x}^{3} + {x}^{2} + kx + 3 - 21 \\ q(x) = \frac{ {x}^{3} + {x}^{2} + kx - 18}{x - 3} \\ q(x) = {x}^{2} + 5x + k + 15 + \frac{ - 18 + 3(15 + k)}{x - 3} \\ q(x) = {x}^{2} + 5x + k + 15 + \frac{3k + 27}{x - 3}$

since x-3 is a factor of f(x),

$3k + 27 = 0 \\ 3k = - 27 \\ k = \frac{ - 27}{3} \\ k = - 9$
Now ,
$q(x) = {x}^{2} + 5x + ( - 9) + 15 \\ q(x) = {x}^{2} + 5x + 6 \\ q(x) = {x}^{2} + 2x + 3x + 6 \\ q(x) = x(x + 2) + 3(x + 2) \\ q(x) = (x + 2)(x + 3)$
for zeros
$(x + 2)(x + 3) = 0 \\ x = - 2 \: or \: - 3$
So the zeroes of the polynomial are 3,-2,-3.