If #x=2# is a zero of #x^3-(k+1)x+k# then what is the value of #k# ?

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Answer 1 · 2017-10-18T18:47:08

#k=6#

Substitute #x=2# into the given cubic equation to get:

#0 = 2^3-(k+1)(2)+k#

#color(white)(0) = 8-2k-2+k#

#color(white)(0) = 6-k#

So:

#k = 6#

With #k = 6# and #x = 2# a zero, #(x-2)# is a factor and we have:

#x^3-(k+1)x+k = x^3-7x+6#

#color(white)(x^3-(k+1)x+k) = (x-2)(x^2+2x-3)#

#color(white)(x^3-(k+1)x+k) = (x-2)(x-1)(x+3)#

So the other two roots are #x=1# and #x=-3#