How do you find k such that #f(x) = x^4 - kx^3 + kx^2 + 1# has the factor x + 2?

1 Answer
Apr 14, 2016

There is no Real value for #k# such that #x+2# is a factor of #f(x)=x^4-kx^3+kx^2+1#

Explanation:

If #(x+2)# is a factor of #f(x)=x^4-kx^3+kx^2+1#
then
#color(white)("XXX")f(x)=(x+2)(g(x))# for some polynomial #g(x)#
from which it is clear that if #x=(-2)# then #f(x)=0#

Substituting #(-2)# for #x# and noting that the result is equal to zero:
#color(white)("XXX")(-2)^2-(-2k)^3+(-2k)^2+1=0#

#color(white)("XXX")16k^4+8k^3+4k^2+1=0#

At (local) minimum/maximum points the slope of
#color(white)("XXX")h(k)=18k^4+8k^3+4k^2+1# must be equal to zero.
Looking for values of #k# such that
#color(white)("XXX")(d(h(k)))/(dk) = 64k^3+24k^2+8k=0#

#color(white)("XXX")8(k)(8k^2+3k+8)=0#

#color(white)("XXX")rArr k=0 or (8k^2+3k+1)=0#
#color(white)("XXX")#...but applying the quadratic formula we can see that
#color(white)("XXXXXX")(k^2+3k+1)=0# has no Real solutions.
So the only critical point occurs when #k=0#
and #f(x)# has a minimum value when #k=0#

Since #f(x) > 0# for all Real values of #x# when #k=0#
#color(white)("XXX")#(since, with #k=0#, #f(x)=x^4+1#)

Therefore there is no value of #k# for which #(x+2)# is a factor of the given expression.