Solve Factoring Problems

Therefore, we were able to create two equations and determine two solutions from this observation.We are left with three factors: 5x, (x 3), and (x - 3). " section, at least one of the three factors must be equal to zero.And making that assumption would cause us to lose half of our solution to this equation. I'll apply the difference-of-squares formula that I've memorized: You can use the Mathway widget below to practice solving quadratic equations by factoring.

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The above, where I showed my checks, is all they're wanting. By the way, you can use this "checking" technique to verify your answers to any "solving" exercise.

So, for instance, if you're not sure of your answer to a "factor and solve" question on the next test, try plugging your answers into the original equation, and confirming that your solutions lead to true statements.

This equation is not in "(quadratic) equals (zero)" form, so I can't try to solve it yet.

The first thing I need to do is get all the terms over on one side, with zero on the other side.

These methods work well for equations like x 2 = 10 - 2x and 2(x - 4) = 0.

But what about equations where the variable carries an exponent, like x Solving the first subproblem, x - 2 = 0, gives x = 2. You should observe that as long as a does not equal 0, b must be equal to zero.

Put another way, the only way for us to get zero when we multiply two (or more) factors together is for one of the factors to have been zero.

So, if we multiply two (or more) factors and get a zero result, then we know that at least one of the factors was itself equal to zero.

In particular, we can set each of the factors equal to zero, and solve the resulting equation for one solution of the original equation.

We can only draw the helpful conclusion about the factors (namely, that one of those factors must have been equal to zero, so we can set the factors equal to zero) if the product itself equals zero.


Comments Solve Factoring Problems

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