Question 1050396
The idea is to plug in x = 2 and replace f(2) with 0. Then solve for k.



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{{{f(x)=x^2+3kx+2k^2}}} Start with the given function



{{{f(2)=(2)^2+3k*(2)+2k^2}}} Replace every x with 2



{{{f(2)=4+6k+2k^2}}} Simplify



{{{0=4+6k+2k^2}}} Replace the "f(2)" with 0



{{{0=2k^2+6k+4}}} Rearrange terms



{{{2k^2+6k+4=0}}} Flip the equation



{{{2(k^2+3k+2)=0}}} Factor out the GCF 2



{{{(2(k^2+3k+2))/2=0/2}}} Divide both sides by 2



{{{k^2+3k+2=0}}} Simplify



{{{(k+1)(k+2)=0}}} Factor the left side



{{{k+1=0}}} or {{{k+2=0}}} Use the zero product property. Then isolate k for each equation.



{{{k+1-1=0-1}}} or {{{k+2-2=0-2}}}



{{{k=-1}}} or {{{k=-2}}}



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So the values of k are {{{k=-1}}} or {{{k=-2}}}