First slide
Theory of expressions
Question

The value of k , so that x43x3+5x233x+k  is divisible by x25x+6  is

Easy
Solution

Let the given equation is f(x)=x43x3+5x233x+k= 0 ….. ( 1 )

Eq. ( 1 ) is divisible by x25x+6 

The factors of given x25x+6  are

(x2)(x3)=0

x=2,x=3

Put x=2  ( or ) x=3  in Eq. ( 1 )

f(2)=243(2)3+5(2)233(2)+k=0

k=54

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