Find k if (#x^3+6x^2+kx+12#) is divisible by (#(x+4)#)?

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Answer 1 · 2017-04-01T17:59:00

THe value of #k=11#

We apply the remainder theorem

Let #f(x)=x^3+6x^2+kx+12#

If #f(x)# is divisible by #(x+4)#, then

#f(-4)=0#

#f(-4)=-64+96-4k+12=0#

#4k=108-64=44#

#k=44/4=11#