Posted by -Untamed- on Thursday, November 24, 2011 at 9:09pm.
Factoring is the easy way, if possible. Not all quadratics can be easily factored, but it is an important method because
If a*b*c*d = 0 then one of a,b,c,d must be zero! That's not true if you set the product to any other number. So, we always set up the equation so that we have a product of factors equal to zero.
In the example you gave, we have
2x(x+3) = 0
so, either
2=0 -- not gonna happen
x=0 -- that's one solution
(x+3)=0 -- so x = -3 is the other solution.
For the other question, we have
x^3 + 8x^2 = 20x
First step: always set things equal to zero
x^3 + 8x^2 - 20x = 0
Now you can see that the x factors out, leaving you with a quadratic:
x(x^2 + 8x - 20) = 0
That's easy to factor, since 10(-2) = -20 and 10 + (-2) = 8
x(x+10)(x-2) = 0
Now we see that either
x=0 -- that's one solution
(x+10)=0 so x = -10 is another
(x-2)=0 so x=2 is the other solution
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