How do you factor #x^3= -3x^2-2x#?

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Answer 1 · 2015-10-13T18:10:53

#x * (x + 1) * (x + 2) = 0#

#x^3 = -3x^2 - 2x#

#iff x * (x^2 + 3x + 2)=0#

Now, choose two numbers, whose sum equals the the coefficient of #x# and whose product is the product of the coefficient of #x^2# and the constant.

Here the coefficient of #x# is #3#
The coefficient of #x^2# is #1#
and the constant is #2#

So the numbers are 2 & 1

Thus the above expression can be written as

#x * (x^2 + 2 x + x + 2)=0#

that is #x * {x * (x + 2) + 1 * (x + 2)}=0#

which in turn can be written as

#x * (x + 1) * (x + 2) = 0#