Question 1089948
<pre><font size=5><b>
2x³+10x²+6x-18=0 

2(x³+5x²+3x-9)=0

Observe that x=1 is a solution to that equation since 

2(1³+5&#8729;1²+3&#8729;1-9) = 2(1+6+3-9) = 0

Therefore x-1 must be a factor of x³+5x²+3x-9.
So we divide by long division:

   <u>     x²+6x+9</u>
x-1)x³+5x²+3x-9
    <u>x³- x²</u>
       6x²+3x
       <u>6x²-6x</u>
           9x-9
           <u>9x-9</u>
              0

As we expect, the remainder is 0, thus the partial
factorization of is x³+5x²+3x-9 is this:   

(x-1)(x²+6x+9) 

and its complete factorization is

(x-1)(x+3)(x+3) or

(x-1)(x+3)²

Therefore the complete factorization of

2x³+10x²+6x-18=0

is

2(x-1)(x+3)² = 0

Edwin</pre></font></b>