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Factoring Complicated Expressions  Problem 1
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!
So factoring a polynomial of higher degree, so what we have here is a polynomial where we we’re dealing with x to the sixth and x to the third. Okay we’re not really used to dealing with anything other than x² and x, but what I want to talk to you about is how these are actually almost exactly the same problem.
If this was x² minus 7x plus 6, we would know how to factor right? Think about the factor of 6 factored up and we’d be okay, but we’re not dealing with x², we’re dealing with x to the sixth. What you need to notice is that the term in the middle is always, the degree on the term in the middle is always half of this term on the end. If we have something of that form, we’re going to be able to factor it the same as something like this, this trinomial we’re used to. This holds that form. Here we’re left with x is 6, x³ is half of that, so they’re really almost identical.
What we have to do then is just make a substitution for that middle term. Let u equal x to the third. So if we’re dealing with this then we’re left with minus 7u plus 6. We still need to figure out what x to the sixth is and what we have is x³, x to the sixth. What do we have to do to x to the sixth, sorry x to the third to get x to the sixth? We have to square it, if we have power to a power we multiply, so x cubed squared is x to the sixth. X³ is actually u, so what this ends up being is just u², so u is x³, x³, x cubed squared is x to the sixth, u is x to the third and what we have is just a different way of writing this exact same equation.
We now have a standard quadratic equation just like we’re talking up here that we know how to factor. I know my leading terms have to be u and u. I know they both have to be negative and factor those 6 are 1 and 6 or 2 or 3, they have to be 1 and 6.
We can’t leave it like this because our equation started with xs, has to end with xs, so take our substitution, plug it back in x³ minus 6, x³ minus 1. So using a fairly simple substitution x to the third which is just that middle term, we are able to factor this down fairly easily.
This one actually has one more step. If you’re told to factor completely that typically means factor it out as much as you possibly can. Here we’re actually dealing with the difference of cubes, so I have a formula for that as well, so we have to finish this up leaving us with x to the third minus 6 doesn’t change, but then we’re dealing with the difference of cubes of x and 1, so we’re left with x minus 1, x² plus x plus 1.
So using the substitution we get down to a factored form which has a difference of cubes so we have to finish it up by using our difference of cubes formula. Factoring using a fairly simple 'u' substitution.
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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.
i love you you are the best, ive spent 3 hours trying to understand probability and this is making sense now finally”
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Factoring Complicated Expressions
Problem 1 5,075 viewsFactor:
x⁶ − 7x³ + 6 
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Problem 2 4,113 viewsFactor:
(3x − 1)² − (x + 3)² 
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Problem 3 3,575 viewsFactor:
6x⁻² − x⁻³ − 2x⁻⁴ 
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Problem 4 3,364 viewsFactor:
x⁶ − 64 
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Problem 5 317 views 
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Problem 6 306 views 
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Problem 7 314 views
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