Question

Using the Remainder Theorem find the remainders obtained when ${\displaystyle x^3+(kx+8)x+k}$ is divided by x + 1 and x - 2 . 

Hence find k if the sum of the two remainders is 1.

Answer 1

Remainder theorem :
Dividend = Divisors ×  Quotient + Remainder

∴ Let f  ( x ) = ${\displaystyle x^3+(kx+8)x+k}$

                 ${\displaystyle =x^3 +k{x}^2+8x+k}$

Dividing f (x ) by x + 1 gives remainder as  R₁

∴  f (-1) = R₁ 

Also , f ( 2 ) = R₂ 

∴  f (-1) = (-1)³ + k (-1)² + 8 (-1) + k

             = -1 + k - 8 + k

             =${\displaystyle 2k-9=R_1}$

${\displaystyle f\left(2\right)={\left(2\right)}^3+k{\left(2\right)}^2+8\times 2+k}$

       ${\displaystyle =8+4k+16+k}$ 

       ${\displaystyle =5k+24=R_2}$

Also,Sum of remainders = ${\displaystyle R_1+ R_1=1}$

∴ ( 2k - 9 ) + ( 5k +24 ) =1 

7k + 15 = 1

7k = -14 

k = - 2