How do you find all the zeros of #h(x)=x^4+6x^3+10x^2+6x+9#?

1 Answer
May 30, 2016

Answer:

#x=-3# (multiplicity 2)

#x=+-i#

Explanation:

#h(x) = x^4+6x^3+10x^2+6x+9#

By the rational root theorem, any rational zeros of #h(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #9# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-3#, #+-9#

In addition, note that all of the coefficients of #h(x)# are positive. So it has no positive zeros. So the only possible rational zeros are:

#-1#, #-3#, #-9#

We find:

#h(-1) = 1-6+10-6+9 = 8#

#h(-3) = 81-162+90-18+9 = 0#

So #x=-3# is a zero and #(x+3)# a factor:

#x^4+6x^3+10x^2+6x+9 = (x+3)(x^3+3x^2+x+3)#

The remaining cubic factors by grouping:

#x^3+3x^2+x+3#

#=(x^3+3x^2)+(x+3)#

#=x^2(x+3)+1(x+3)#

#=(x^2+1)(x+3)#

#=(x-i)(x+i)(x+3)#

So #x=-3# is a zero again, giving it a total multiplicity #2# and the other zeros are #x=+-i#