Question 1154821
<pre>

{{{2x^4 + 13x^3 + 22x^2 + 8x = 0}}}

We can factor out x

{{{x(2x^3 + 13x^2 + 22x + 8) = 0}}}

Since -2 is one of its zeros, the polynomial in parentheses must be divisible by
(x+2) because (x+2) would have to be one of the factors that you'd have to set
equal to zero in order to get x = -2.  We'll use synthetic division

-2|2 13  22  8  
  |<u>  -4 -18 -8</u>  
   2  9   4  0  

So we have factored the left side of the polynomial equation as

{{{x(x+2)(2x^2 + 9x + 4) = 0}}}

We can factor further as

{{{x(x+2)(x+4)(2x+1)=0}}}

Use the zero-factor property:

x=0;   x+2=0;    x+4=0;     2x+1=0
         x=-2      x=-4;      2x=-1
                               x=-1/2

The four zeros are 0; -2; -4; -1/2

Edwin</pre>