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Factoring Complicated Expressions  Problem 4
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Factoring an expression with only two terms; so whenever we have an expression with two terms, we really have two options that we’re dealing with. We’re dealing with difference of squares or the difference or the sum of cubes okay? Obviously we have a negative here, so we’re going to be dealing with the difference of one of those two things, squares, or cubes.
For this particular problem, we actually have options, we can do it either way because 64 is both the square and the cube and x to the 6 is also both the square and the cube so you have choices. I’m going to do it as the difference of cubes. You can do it as the difference of squares as well you’ll get the same answer.
So what we want to do is figure out how we can relate this to a³ minus b³ because we have a formula for how this factors. 64 is 4³ that’s easy enough and we want to figure out what something else is cubed to give us x to the sixth. We know there has to be an x involved and we know that there probably has to be a power on our x. When we take a power to a power and multiply, so we want to figure out what times 3 is equal to 6. This is going to be x². So we know that we have x² to the third minus 4 to the third as our difference of cubes. From here we just have our formula.
This is just going to be a minus b times a² plus ab plus b². So all we really have to do then is plug these terms in in order to make it work okay? So what we have is, let’s get a different color up here. X² minus 4 times a², so x², squared is x to the fourth. The opposite sign, a times be so this turns into 4x² and then b² is just going to be plus 16. For this particular one what we’re actually left with is an x² minus 4 which is the difference of squares, so we can take this one more step leaving us with x minus 2, x plus 2, x to the fourth plus 4x² plus 16.
So by making a substitution by seeing what these are cubed, you’re able to factor this out. If you did it as a difference of squares, your answer actually might look slightly different, that’s okay, you’ve still factored out, you’ve still done everything you can, what happens in this case is we’re not able to factor this you do it the other way you’re going to get a different point where something else isn’t going to factor quite right either.
So equation like this you could do difference of squares, or cubes it doesn’t matter which just pick one and go with it.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Sample Problems (7)
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Factoring Complicated Expressions
Problem 1 5,503 viewsFactor:
x⁶ − 7x³ + 6 
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Problem 2 4,491 viewsFactor:
(3x − 1)² − (x + 3)² 
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Problem 3 3,879 viewsFactor:
6x⁻² − x⁻³ − 2x⁻⁴ 
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Problem 4 3,660 viewsFactor:
x⁶ − 64 
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Problem 5 496 views 
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Problem 6 482 views 
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Problem 7 463 views
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