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

##### 1 Answer

#### Explanation:

By the rational root theorem, any rational zeros of

That means that the only possible rational zeros are:

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

In addition, note that all of the coefficients of

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

We find:

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

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

So

#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