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

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This question is very tricky.

its an important 5 mark question for boards exam

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1. Given, p(x) = x³ + 2x² + kx +3

g(x) = x – 3

Remainder, r(x=3) = 21.

We have to find the quotient and the value of k and the zeros of the cubic polynomial.

The division algorithm states that given any polynomial p(x) and any non-zero

polynomial g(x), there are polynomials q(x) and r(x) such that

p(x) = g(x) q(x) + r(x)

To find the value of k, put x = 3 in p(x),

p(3) = (3)³ + 2(3)² + k(3) +3 = 21

27 + 18 + 3k + 3 = 21

48 + 3k = 21

3k = -27

k = -27/3

k = -9

Therefore, the cubic polynomial is x³ + 2x² – 9x + 3.

Using long division to find the quotient,

The quotient is x² + 5x + 6.

On factoring,

x² + 5x + 6 = 0

x² +3x + 2x + 6 = 0

x(x + 3) + 2(x + 3) = 0

(x + 2)(x + 3) = 0

Now, x + 2 = 0

x = -2

Also, x + 3 = 0

x = -3

Therefore, the zeros of the polynomial are 3, -2 and -3

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