How do you find all the zeros of #f(x) = x^3 + 11x^2 + 41x + 51#?

1 Answer
Mar 21, 2016

Answer:

Use the rational root theorem to find zero #x = -3#, divide by #(x+3)# then solve the remaining quadratic to find zeros #x = -4+-i#.

Explanation:

#f(x) = x^3+11x^2+41x+51#

by the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p# and #q# where #p# is a divisor of the constant term #51# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1#, #+-3#, #+-17#, #+-51#

In addition, note that all of the coefficients of #f(x)# are positive, so it can only have zeros for negative values of #x#.

That leaves:

#-1#, #-3#, #-17#, #-51#

Trying each in turn we find:

#f(-1) = -1+11-41+51 = 20#

#f(-3) = -27+99-123+51 = 0#

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

#x^3+11x^2+41x+51 = (x+3)(x^2+8x+17)#

Then solve:

#0 = x^2+8x+17#

#= x^2+8x+16+1#

#= (x+4)^2-i^2#

#= ((x+4)-i)((x+4)+i)#

#= (x+4-i)(x+4+i)#

So the remaining zeros are: #x = -4+-i#