SOLUTION: Solve using the principle of zero products (x - 2)(2x + 8)(3x + 9) = 0 this is what i cam up with am i doing this right? 30 + -21x + 3x2 = 0 3(10 + -7x + x^2) = 0 gcf is 3 3(

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Solve using the principle of zero products (x - 2)(2x + 8)(3x + 9) = 0 this is what i cam up with am i doing this right? 30 + -21x + 3x2 = 0 3(10 + -7x + x^2) = 0 gcf is 3 3(      Log On


   

Question 325625: Solve using the principle of zero products
(x - 2)(2x + 8)(3x + 9) = 0
this is what i cam up with am i doing this right?
30 + -21x + 3x2 = 0
3(10 + -7x + x^2) = 0 gcf is 3
3((2 + -1x)(5 + -1x)) = 0
2 + -1x = 0
2 + -2 + -1x = 0 + -2
0 + -2 = -2
-1x = -2
Divide each side by '-1'.
x = 2
5 + -1x = 0
5 + -5 + -1x = 0 + -5
5 + -5 = 0
0 + -1x = 0 + -5
-1x = 0 + -5
0 + -5 = -5
-1x = -5
0 + -5 = -5
-1x = -5
Divide each side by '-1'.
x = 5
x = {2, 5} am i doing this right????
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
%28x+-+2%29%282x+%2B+8%29%283x+%2B+9%29+=+0
a 2 and 3 can be factored out
%28x+-+2%292%28x+%2B+4%293%28x+%2B+3%29+=+0
6%28x+-+2%29%28x+%2B+4%29%28x+%2B+3%29+=+0
set each = to 0
6%28x+-+2%29=0
highlight%28x=2%29
%28x+%2B+4%29=0
highlight%28x=-4%29
%28x+%2B+3%29=0
highlight%28x=-3%29
x={-4,-3,2}