Solving a Quadratic Equation by Factoring Question

May 2010
22
0
I have the problem (x+2)^3 = x^3+8
I use the cube formula (x+y)^(3) = x^(3) + 3x^(2)y + 3xy^(2) + y^(3) and get 6x^(2) + 12x = 0 and solve for x and get -2.

But the book says 0 is also an answer. Can someone please tell me why, and why sometimes 0 is not an answer?

Thanks.
 

skeeter

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I have the problem (x+2)^3 = x^3+8
I use the cube formula (x+y)^(3) = x^(3) + 3x^(2)y + 3xy^(2) + y^(3) and get 6x^(2) + 12x = 0 and solve for x and get -2.

But the book says 0 is also an answer. Can someone please tell me why, and why sometimes 0 is not an answer?

Thanks.
\(\displaystyle 6x^2 + 12x = 0
\)

factor ...

\(\displaystyle 6x(x + 2) = 0\)

set each factor equal to 0 ...

\(\displaystyle 6x = 0\)

\(\displaystyle x+2 = 0\)
 
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May 2010
22
0
Thanks

I forgot to factor before setting it equal to zero. Btw, how do you make your math text like that?