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### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 1

Lesson 2: Topic B: Lessons 12-13: Factorization- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Factor polynomials: quadratic methods
- Factoring two-variable quadratics
- Factoring two-variable quadratics: rearranging
- Factoring two-variable quadratics: grouping
- Factor polynomials: quadratic methods (challenge)
- Factoring using the perfect square pattern
- Factoring using the difference of squares pattern
- Factoring difference of squares: two variables (example 2)
- Factor polynomials using structure
- Factoring higher-degree polynomials
- Factoring sum of cubes
- Factoring difference of cubes
- Finding zeros of polynomials (1 of 2)
- Finding zeros of polynomials (2 of 2)
- Finding zeros of polynomials (example 2)
- Zeros of polynomials (with factoring)

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# Factoring difference of cubes

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.

## Want to join the conversation?

- 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)

- 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.(2 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.(4 votes)

- 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)

- Why is factoring cubes so difficult? :((1 vote)
- You need to learn the pattern. It is confusing initially until you understand how to use the pattern and have it memorized.(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)

## Video transcript

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.