Question

7x^3+56

factor expression completely.

Answer 1

Remember what I told you before, first always try to factor a common factor.
Do 7x^3 and 56 have a common factor?

Answer 2

3?

Answer 3

No. Neither 7 nor 56 is divisible by 3. What is 7 divisible by?

Answer 4

7

Answer 5

or 1

Answer 6

Right. Is 56 also divisible by 7?

Answer 7

yes

Answer 8

Both terms are divisible by 7. That's the common factor.
Now, factor out the common factor of 7.

Answer 9

..1?

Answer 10

7x^3 + 56 =
7(x^3 + 8)
Factoring is the opposite of multiplying. It's undoing a multiplication.

Answer 11

Now that you see that
7x^3 + 56 is equal to
7(x^3 + 8), what does x^3 + 8 look like to you (of the types of factorizations you've been doing)?

Answer 12

so dividing it by 7 made it 8?

Answer 13

Yes, the second term of 56 divided by 7 equals 8.

Answer 14

okay so now what

Answer 15

@mathstudent55

Answer 16

Now you have 7(x^3 + 8)
Does the part (x^3 + 8) look like a type of factoring you've just done?
x^3 is the cube of x.
8 is the cube of 2.
Does this look familiar?

Answer 17

so its prime

Answer 18

or not

Answer 19

my bad its not

Answer 20

No, it's a sum of cubes. You just did z^3 + 27, which is also a sum of cubes.

Answer 21

but i still didnt get that one. that answer was wrong

Answer 22

Yes, you are right. That answer had an error. I fixed it now.

Answer 23

The factorization of
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
Here you have x^3 + 8, so a = x and b = 2
x^3 + 8 = (x + 2)(x^2 - 2x + 4)

Answer 24

Just keep in mind that this problem did not start as x^3 + 8. It started as 7x^3 + 56, so the complete factorizxation is:
7x^3 + 56 = 7(x + 2)(x^2 - 2x + 4)