What are the values of #k# when #x^4-3x^2+x-2# is divided by #x-k# and has a remainder of #2#?

1 Answer
May 6, 2018

#k=-2.094945# and #1.893974#

Explanation:

According to remainder theorem , when we divide a polynomial #f(x)# by #x-k# remainder is #f(k)#.

Therefore dividing #x^4-3x^2+x-2# by #x-k# we get

#k^4-3k^2+k-2# and as remainder is #2#, we have

#k^4-3k^2+k-2=2# or #k^4-3k^2+k-4=0#

Further according to rational zeros theorem , If #f(x)# is a polynomial with integer coefficients and if #p/q# is a zero of #f(x)# i.e. #f(p/q)=0#, then #p# is a factor of the constant term of #f(x)# and #q# is a factor of the leading coefficient of #f(x)#.

The factors of #-4# are #+-1,+-2,+-4# and as for none of them as #k#, #k^4-3k^2+k-4=0#, we do not have rational roots.

The graph appears like this and solutions are slightly less than #-2# and #2#. These values obtained using Goal Seek in MS Excel are #-2.094945# and #1.893974#.

graph{x^4-3x^2+x-4 [-3, 3, -5, 5]}