Question

the polynomial p(x)=kx³ + qx² + 4x - 8 when divided by (x+3) leaves a remainder 20 (1-k). Find the value of a

Answer 1

p(x) = kx^3 + qx^2 + 4x - 8
when it is divided by x + 3, it leaves a remainder 20(1-k)
=> (x+3) * (some value) + 20(1-k) = p(x) = kx^3 + qx^2 +4x - 8
=> (x+3) * (some value) = kx^3 + qx^2 + 4x - 8 - 20(1-k)
=> (x+3) * (some value) = kx^3 + qx^2 + 4x - 8 - 20 + 20k
let rhs be r(x)
=> (x+3) does not leave any remainder for  r(x)
=> (x+3) is a root of r(x)
=> When x = -3 , r(x) = 0
=> k(-3)^3 + q(-3)^2 +4(-3) - 8 -20 + 20k = 0
=> -27k + 9q -12 -28 +20k = 0
=> -7k + 9q -40 = 0
=> 9q = 7k + 40
=> q = (7k+40)/9