WEBVTT
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Find the solution set of two π₯ cubed equals 32π₯ in the real numbers.
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First, we should get our equation equal to zero.
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And we want to keep the variable with the highest exponent positive, so π₯ cubed β that needs to stay positive.
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So the two π₯ cubed will stay positive.
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So we need to subtract the 32π₯ from both sides of the equation, which leaves us with two π₯ cubed minus 32π₯ equals zero.
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So since there are two terms, we can factor by first taking out a greatest common factor.
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And we can take out a two and an π₯.
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If we take out two π₯ from two π₯ cubed, weβre left with π₯ squared.
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If we take out two π₯ from 32π₯, weβre left with 16.
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Now, inside the parentheses, we have a difference of two squares because π₯ squared we can square root and 16 we can square root.
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The formula for difference of squares is π squared minus π squared equals π plus π times π minus π.
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So if we have π₯ squared minus 16, the square root of π₯ squared is π₯ and the square root of 16 is four.
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So we will have π₯ plus four and π₯ minus four.
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So now, weβve completely factored our equation.
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So to find the solution, we need to take each factor and set it equal to zero.
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So we set two π₯ equal to zero, we set π₯ plus four equal to zero, and we set π₯ minus four equal to zero.
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So for two π₯ equals zero, we need to divide both sides of the equation by two.
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And we get that π₯ is equal to zero.
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For π₯ plus four equals zero, we need to subtract four from both sides of the equation.
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And weβre left with π₯ equals negative four.
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For π₯ minus four equals zero, we need to add four to both sides of the equation.
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And we have π₯ equals four.
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So our solution set would be zero, four, and negative four.
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And the order does not matter.