Sal factors 40c^3-5d^3 as 5(2x-d)(4c^2+2cd+d^2) using a special product form for a difference of cubes. Created by Sal Khan.
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- How to solve the equation:x^3-6x^2+9x=4? The x-values when y is 4 is 1 and 4. I don't know how to get that.(5 votes)
- how to prove the accuracy of the formula for difference of cube?(1 vote)
- So we have a^3 - b^3 = (a-b)(a^2 + ab + b^2)
If you simply multiply out these brackets you will get a^3 - b^3, so it is accurate.(7 votes)
- How would you factor 5x^3-40 ? and 4x^10-y^6?(2 votes)
- These seem like homework problems which you are suppose to do. So, I'll give you some hints.
1) 5x^3-40: This polynomial has a common factor. Factor it out as your 1st step. Then, the new binomial will be a difference of cubes. Factor it using the techniques shown in this video.
2) 4x^10-y^6: This polynomial is the difference of 2 squares. Here's a link to the video covering that topic: https://www.khanacademy.org/math/algebra2/polynomial-functions/factoring-polynomials-special-product-forms-alg2/v/factoring-difference-of-squares
Hope this helps.(5 votes)
- what is this 5(2c-d)(4c^2+2cd+d^2) for ? and why did we factor it out?(4 votes)
- We factor it because it makes it simpler to find solutions. If we are dealing with the equation 40c^3 - 5d^3 = 0, for example, it is not easy to find the solutions of a cubic equation, so we factorise it into a linear equation and a quadratic, both of which we can find solutions for very easily.(1 vote)
- Okay, so I understand that the rule is, if you get something like
a^3 + b^3,
the factored version is gonna have a '-' between the first and second terms in the trinomial, and if you get
a^3 - b^3,
you'll get a '+' between those terms.
Either way, there's always a '+' between the second and third terms.
What happens though, if you get a problem where both cubed terms are negative?
Does THAT give you a '-' between the second and third terms in the factored form?
(I hope I worded this question okay.)(2 votes)
- If both terms are negative like: -a^3 - b^3
1) Factor out a common factor of -1: -1 (a^3+b^3)
2) then, factor the sum of cubes: -1 (a+b) (a^2-ab+b^2)
Hope this helps.(2 votes)
- I am confused. I know that the video says that "d" is the b value, but couldn't you use "-d" as your b value in the sum of cubes? Or when you do the difference of cubes is it supposed to be "-d" Help(2 votes)
- Either way would be fine. Difference of cubes is not really a separate rule, but it can seem a bit confusing since difference of squares and sum of squares are different rules. But your intuition is correct. You could write the sample polynomial as 5((2c)^3 + (-d)^3)(2 votes)
- i have sort of a similar problem that asks me to factor COMPLETELY
x^6 - y^6 how to would i completely factor this after using the differences of cubes(1 vote)
- Actually, you can go one more step, I believe. You can then use the difference of cubes for (x^3-y^3), and can use sum of cubes on (x^3+y^3). Now that I think about it, it may not be getting any further...
You should get (x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)(2 votes)
- Could you fist simplify it to -5((-2c)^3+d^3) and factor it from there or is one way better?(1 vote)
- There is no need to factor out a -5. Usually you would factor out a negative when the lead coefficient is negative. That is not the case here.
You have changed the difference of 2 cubes into a sum of 2 cubes. Both are factorable, but use slightly different patterns.(2 votes)
- Shouldn’t it be factored into 5(2c+d)(4c^2-2cd+d^2)? It is a negative d so when you put it into the difference of cubes formula, the negatives would cancel out.(1 vote)
- You can check your factors by multiplying them. They do not create 40c^3-5d^3. Instead, they create: 40c^3+5d^3. The sign in the middle is different.(2 votes)
- Where on Khan Academy can I find practice problems for factoring difference of cubes?(1 vote)
- Try the search bar that's at the top of all KA screens and see if they are in a difference section.(2 votes)
We're asked to factor 40c to the third power minus 5d to the third power. So the first thing that might jump out at you is that 5 is a factor of both of these terms. I could rewrite this as 5 times 8c to the third power minus 5 times d to the third power. And so you could actually factor out a 5 here, so factor out a 5. And so if you factor out a 5, you get 5 times 8c to the third power minus d to the third power. So as you see, factoring, it really is just undistributing the 5, reversing the distributive property. And when you write it like this, it might jump out at you that 8 is a perfect cube. It's 2 to the third power. c to the third power is obviously c to the third power. And then you have d to the third power. So this right here is a difference of cubes. And actually, let me write that explicitly because 8 is the same thing as 2 to the third power. So you can write this as-- let me write the 5 out front-- 5 times-- this term right over here can be rewritten as 2c to the third power because it's 2 to the third power times c to the third power-- 8c to the third power-- And then, minus d to the third power. And so this gives us, right over here, a difference of cubes. And you can actually factor a difference of cubes. And you may or may not know the pattern. So if I have a to the third minus b to the third, this can be factored as a minus b times a squared plus ab plus b squared. And if you don't believe me, I encourage you to multiply this out, and you will get a to the third minus b to the third. You get a bunch of terms that cancel out, so you're only left with two terms. And even though it's not applicable here, it's also good to know that the sum of cubes is also factorable. It's factorable as a plus b times a squared minus ab plus b squared. So once again, I won't go through the time of multiplying this out, but I encourage you to do so. It just takes a little bit of polynomial multiplication. And you'll be able to prove to yourself that this is indeed the case. Now, assuming that this is the case, we can just do a little bit of pattern matching. Because in this case, our a is 2c, and our b is d. So let me write this. a is equal to 2c, and our b is equal to d. We have minus b to the third and minus d to the third, so b and d must be the same thing. So this part inside must factor out to-- let me write my 5, open parentheses. Let me give myself some space. So it's going to factor out into a minus b. So a is 2c minus b, which is d. So it factors out as the difference of the two things that I'm taking the cube of. 2c minus d times-- and now, I have a squared is 2c squared. 2c squared is the same thing as 4c squared. Let me make that. a squared is equal to 2c-- the whole thing squared, which is equal to 4c squared. So it's 4c squared plus a times b. So that's going to be 2c times d, so plus 2c times d. And then finally plus b squared, and in our case, b is d. So you get plus d squared. And you're done. We have factored it out. And actually, you could get rid of one set of parentheses. This can be factored as 5 times 2c minus d times 4c squared plus 2cd plus d squared. And we are done.