6x^6-48 factor completely

Question

anonymous · Answer

im sorry what was that

campbell_st · Answer

factor out 6 and you are left with 

$$6(x^6 - 8)$$

make a substitution 

$$u = x^2 $$

and you have 

$$6(u^3 - 8)$$

this is the difference of 2 cubes and can be factored

hope it helps

anonymous · Answer

is it not factor
bc it would be -6 + x6 =48

campbell_st · Answer

just read the solution I posted

anonymous · Answer

6(u^3)-4 i do not know if this is right

campbell_st · Answer

read the solution and all you need to do is factor the difference of 2 cubes

anonymous · Answer

i do not know how

mathstudent55 · Answer

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

campbell_st · Answer

6 multiplied by -8 = -48

that seems to be a mistake you have made

anonymous · Answer

im sorry

anonymous · Answer

6 (u)-8

anonymous · Answer

can you show me step by step

anonymous · Answer

?

jhannybean · Answer

What @campbell_st posted was if you make a substitution for x$^6$ as u = x$^2$, then you can think of your factored function as: $$6(\color{red}{(x^2)}^3-2^3)$$ which is the difference of cubes.

Difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

All you have to do is define $a=x^2$  and $b=2$ and fill in the form

mathstudent55 · Answer

You need to use the pattern I wrote above for factoring a difference of cubes.

jhannybean · Answer

Forgot to tag @mathstudent55 for the nicely written formula posted above ^^

jhannybean · Answer

Forgot to tag @mathstudent55 for the nicely written formula posted above ^^

mathstudent55 · Answer

@anthony6577 
If you will pay attention, I will help you from beginning to end with this problem.

anonymous · Answer

im trying

mathstudent55 · Answer

Ok, here we go.

Factor: $\large 6x^6 - 48$

The first step in factoring is always to try to factor a common factor.

anonymous · Answer

alright

mathstudent55 · Answer

Here, there is a common factor in both terms. that common factor is 6, so we factor it out.

$\large = 6(x^6 - 8)$

mathstudent55 · Answer

The 6 will remain as a factor to the end, but we don;t need to worry about it.
We need to figure out what to do with x^6 - 8.

anonymous · Answer

so the gcf is 6

mathstudent55 · Answer

To factor a binomial, there are 3 possibilities:
1. A difference of squares.
2. A sum of cubes.
3. A difference of cubes.

mathstudent55 · Answer

We look carefully at x^6 - 8 to figure out which of the three possibilities above, if any, it is.

anonymous · Answer

diffrent of cubs

mathstudent55 · Answer

We notice that x^6 can be broken up in two ways:

$\large x^6 = (x^3)^2$ in this case, x^6 is the square of x^3, and is a square.

$\large x^2 = (x^2)^3 $ in this case, x^6 is the cube of x^2 and is a cube.

mathstudent55 · Answer

We don't know yet whether we are dealing with cubes or squares.
Now we look at 8.

mathstudent55 · Answer

With 8, we notice that
$\large 8 = 2^3$
8 is indeed a cube, it is the cube of 2.

mathstudent55 · Answer

Since 8 is a cube, we need to treat x^6 as a cube also, so we have a difference of cubes.

mathstudent55 · Answer

Now we know we are dealing with a difference of cubes, so we write x^6 - 8 as a difference of cubes clearly showing the cubes.

anonymous · Answer

so the anser would be 8+2^3

mathstudent55 · Answer

$\large =6((x^2)^3 - 2^3) $

mathstudent55 · Answer

No. We're not there yet.
We now need to use the pattern of factoring a difference of cubes I wrote above.

mathstudent55 · Answer

$\large a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

anonymous · Answer

this pint threw me off

mathstudent55 · Answer

The way a difference of cubes is factored is to first find what the cubes are cubes of.
x^6 is the cube of x^2
8 is the cube of 2

anonymous · Answer

so 2

mathstudent55 · Answer

Now you need two sets of parentheses.
In the first one, you need the difference of the quantities without cubes.
(x^2 - 2)

anonymous · Answer

alright

mathstudent55 · Answer

In the second set of parentheses, you need the square of the first one: (x^2)^2 = x^4
Then add to it the product of the two0: x^2 * 2 = + 2x^2
Then add to that the square of the second one: 2^2 = + 4
Now fill in the second set of parentheses with these 3 quantities:
(x^4 + 2x^2 + 4)

mathstudent55 · Answer

Finally, we can write the complete solution:

$\large 6x^6 - 48$

$\large =6(x^6 - 8)$

$\large = 6((x^2)^3 - 2^3)$

$\large = 6(x^2 - 2)(x^4 + 2x^2 + 4) $

anonymous · Answer

so that is the work

mathstudent55 · Answer

yes, and that is also the final answer

anonymous · Answer

that alot just to find out

mathstudent55 · Answer

The solution is quick, but with all the explanations, it becomes long.

anonymous · Answer

thank you for spending the time to show me

mathstudent55 · Answer

You are welcome.