Question 1007037
<pre>Solve each polynomial equation by factoring. 

1. 

5x³+10x²+5x = 0

Factor out 5x

5x(x²+2x+1) = 0

Factor the quadratic in the parentheses:

5x(x+1)(x+1) = 0

5x=0; x+1=0;  x+1=0
 x=0;   x=-1    x=-1    

</pre>
For #1 - Would this answer be cannot be factored ? 
<pre>
No, for we factored it twice.
</pre>
Because I did it, and i'm getting x = {0, -1, -1}.
<pre>
That's the correct solution but the only way to get it
is to factor the polynomial.  

2. 

x³+2x²-9x-18 = 0

Factor x² out of the first two terms on the left 
and factor -9 out of the last two terms on the left:

x²(x+2)-9(x+2) = 0

We factor out the common factor (x+2)

(x+2)(x²-9) = 0

Factor the expression in the second parentheses as
the difference of two squares:

(x+2)(x-3)(x+3) = 0

Use the zero-factor property:

x+2=0;  x-3=0;  x+3=0
  x=-2;   x=3;    x=-3

Edwin</pre>