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

##### 1 Answer

Use the rational root theorem to find zero

#### Explanation:

by the rational root theorem, any rational zeros of

That means that the only possible rational roots are:

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

In addition, note that all of the coefficients of

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+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: