A x 2 100 0 solution b 25 y 2 3 0 solution c

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Unformatted text preview: found the LCD we also saw that we needed to avoid x = 1 so we didn’t get division by zero. Therefore, this equation has a single solution, x = -4 [Return to Problems] Before proceeding to the next topic we should address that this idea of factoring can be used to solve equations with degree larger than two as well. Consider the following example. Example 3 Solve 5 x3 - 5 x 2 - 10 x = 0 . Solution The first thing to do is factor this equation as much as possible. In this case that means factoring out the greatest common factor first. Here is the factored form of this equation. © 2007 Paul Dawkins 89 http://tutorial.math.lamar.edu/terms.aspx College Algebra 5x ( x2 - x - 2) = 0 5 x ( x - 2 )( x + 1) = 0 Now, the zero factor property will still hold here. In this case we have a product of three terms that is zero. The only way this product can be zero is if one of the terms is zero. This means that, 5x = 0 x-2=0 x +1 = 0 Þ Þ Þ x=0 x=2 x = -1 So, we have three solutions to this equation. So, provided we can factor a polynomial we can always use this a...
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