If the remainder on division x 3 + 2 x 2 + k x + 3 by x - 3 is 21 , find the value of k and hence find the zeroes of the cubic polynomial x 3 + 2 x 2 + k x - 18 .

Given, dividend, x 3 + 2 x 2 + k x + 3 by x - 3 is 21 . Using remainder theorem, we get, f 3 = remainder ⇒ 3 3 + 2 3 2 + 3 k + 3 = 21 ⇒ 27 + 18 + 3 k + 3 = 21 ⇒ 48 + 3 k = 21 ⇒ 3 k = - 27 ⇒ k = - 9 Therefore, the second polynomial is x 3 + 2 x 2 - 9 x - 18 . Thus, x 3 + 2 x 2 - 9 x - 18 = x 2 x + 2 - 9 x + 2 = x 2 - 9 x + 2 = x 2 - 3 2 x + 2 = x + 3 x + 2 x - 3 Therefore, the zeroes of the polynomial are 3 , - 3 and - 2 .

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Find the remainder when $x^{3} + 3 x^{2} + 3 x + 1$ is divided by $x + 1$ .

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Find the remainder when x3+3x2+3x+1 is divided by x+1.

Important Questions on Polynomials

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Find the remainder when $x^{3} + 3 x^{2} + 3 x + 1$ is divided by $x + 1$ .