Question

is the remainder of the division of x cube + 2 X square + 3 + K X by x minus 3 is 21 find the quotient and the value of k hence find the zeros of the cubic polynomial x cube + 2 X square + kx - 18

Answer 1

Correct Question : If  the remainder of the 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.

Solution : We have to subtract remainder from dividend to make it completely divisible.

→ x³ + 2x² + kx + 3 - 21

→ x³ + 2x² + kx - 18

Now on dividing above equation by x - 3, (please refer to attatchment).

→ Remainder = 27 + 3k

But we know that remainder should be zero since we have decreased remainder from it therefore,

⇒ 0 = 27 + 3k

⇒ - 27/3 = k

⇒ k = - 9

By substituting value we get,

→ p(x) = (x - 3)(x² + 5x + 15 - 9)

→ p(x) = (x - 3)(x² + 5x + 6)

→ p(x) = (x - 3)(x² + 3x + 2x + 6)

→ p(x) = (x - 3)(x + 2)(x + 3)

Hence other zeros are ± 3 and - 2.