Question

the remainder on division
$x ^{3} + 2x ^{2} + kx + 3$
by
$x - 3$
is 21, find the quotient and the value of k. hence find the zeros of cubic polynomial
$x ^{3} + 2x ^{2} + kx - 18$

Answer 1

Hi ,

Let p ( x ) = x³ + 2x² + kx + 3 ,

If p( x ) is divided by ( x - 3 ) then the

remainder is p ( 3 ) ,

But according to the problem given ,

p( 3 ) = 21

3³ + 2( 3)² + k × 3 + 3 = 21

27 + 18 + 3k + 3 = 21

27 + 3k = 0

3k = - 27

k = ( -27 ) /3

k = - 9

Now ,

x³ + 2x² + kx -18

= x³ + 2x² - 9x -18 = g ( x )

If x = -2

g ( x ) = ( -2 )³ + 2 ( -2 )² - 9 ( -2 ) - 18

= -8 + 8 +18 - 18

= 0

Therefore ,

( x + 2 ) is a factor of g ( x )

g ( x ) = ( x + 2 ) ( x² - 9 )

= ( x + 2 ) ( x² - 3² )

= ( x + 2 ) ( x + 3 ) ( x - 3 )

Therefore ,

Zeroes of g ( x ) is -2 , - 3 , 3

I hope this helps you.

:)