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Examine the equation below:
ab = 0
If you let a = 3, then logivally b must equal 0. Similarly, if you let b = 10, then a must equal 0.
Now try letting a be some other non-zero number. You should observe that as long as a does not equal 0, b must be equal to zero.
To state the observation more generally, "If ab = 0, then either a = 0 or b = 0." This is an important property of zero which we exploit when solving by factoring.
When the example was factored into (x - 2)(x - 3) = 0, this property was applied to determine that either (x - 2) must equal zero, or (x - 3) must equal zero. Therefore, we were able to create two equations and determine two solutions from this observation.
5x3 = 45x
Move all terms to the left side of the equation. We do this by subtracting 45x from each side.
5x3 - 45x = 45x - 45x
5x3 - 45x = 0.
The next step is to factor the left side completely. We first note that the two terms on the left have a greatest common factor of 5x.
5x(x2 - 9) = 0
Now, (x2 - 9) can be factored as a difference between two squares.
5x(x + 3)(x - 3) = 0
We are left with three factors: 5x, (x + 3), and (x - 3). As explained in the "Why does it work?" section, at least one of the three factors must be equal to zero.
Create three subproblems by setting each factor equal to zero.
1. 5x = 0
2. x + 3 = 0
3. x - 3 = 0
Solving the first equation gives x = 0. Solving the second equation gives x = -3. And solving the third equation gives x= 3.
The final solution is formed from the solutions to the three subproblems.
x = -3, 0, 3
Visit the next page for another example.
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