How do you find all the zeros of #h(x) = x^4 + 10x^3 + 26x^2 + 10x + 25#?

1 Answer
Mar 20, 2016

Answer:

Use the rational root theorem to help find the first zero, then factor by it and by grouping to find the other roots.

Explanation:

#h(x) = x^4+10x^3+26x^2+10x+25#

By the rational root theorem, any rational roots of #h(x)=0# must be expressible in the form #p/q# for integers #p# and #q# where #p# is a factor of the constant term #25# and #q# a factor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-5#, #+-25#

In addition, note that all of the coefficients of #h(x)# are positive. So it can have no positive zeros. So that just leaves the possibilities:

#-1#, #-5#, #-25#

We find:

#h(-1) = 1-10+26-10+25 = 32#

#h(-5) = 625-1250+650-50+25 = 0#

So #x = -5# is a zero and #(x+5)# a factor:

#x^4+10x^3+26x^2+10x+25#

#=(x+5)(x^3+5x^2+x+5)#

#=(x+5)(x^2(x+5)+1(x+5))#

#=(x+5)^2(x^2+1)#

So #x=-5# is a double zero and the other zeros are #x=+-i#