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Question

If x3+8x2+kx+18 is completely divisible by x2+6x+9, then find the value of 'k'.

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Solution

x3+8x2+kx+18 is completely divisible by x2+6x+9

First of ll we should break x2=6x+9

x2+2.3x+x2+(x+3)2

Hence x=3 is the zero of x3+8x2+kx+18

Now put x=3 in x3+8x+kx+18

(3)3+8(3)2+k(3)+18=0

27+8×93k+18=0

27+723k+18=0

633k=0

k=63/3

k=21

k=21

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