Solving Quadratic Equations By Factoring Practice Problems

Solving Quadratic Equations By Factoring Practice Problems-73
We do this by looking for a pair of numbers that have a product equal to the constant on the end of the trinomial and a sum equal to the - 3x - 10 into (x 2)(x - 5) by realizing that 2 * -5 = -10, and 2 -5 = -3. The goal is still the same - split the trinomial into a product of binomials - and we'll still find a lot of the same patterns, but now we'll have to make two slight changes in the process in order to end up with the correct answer.Let's go ahead and take a look at the example I just mentioned.

Factor 2x We'll start this problem very similarly to the simple factoring problems by looking for two numbers that fit the pattern.

The thing is, it's not going to be quite the same pattern.

You can always quickly multiply out your answer to make sure you got the right thing, and if you do that here, looks like we're good. Not only is this example another one with a non-1 leading coefficient, it's also an example of a special quadratic that often messes students up. Now that we have these two values, we can fill in our area model.

We put the 9x and the -4 in one diagonal and the 6x and -6x in the other, but it actually doesn't matter which diagonal or even which order you put them in; they're all going to work!

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Rewriting the terms from each side together in parentheses as binomials says that the factored form of this is (3x 2)(3x - 2).

Factoring problems with a leading coefficient that isn't 1 have two differences from their simpler counterparts.

Putting the 2x If we can find what terms must have been on the outside of this chart to get multiplied in and give us what we have here, we'll be done.

We do this by dividing out the greatest common factor from each row and column of our chart.

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