How do you factor (27x^3+8)?

Question

What are the steps to factoring (27x^3+8)? I know the answer is x(3x+2)(9x^2 - 6x +4). But, how do you get there? Can somebody briefly show me the steps?

Thanks !!

Ok--Thanks!! I totally understand now! I'll choose a best answer when 4 hours is up! Thank you so much!

Answer 1

There are only a few ways to factor a third degree polynomial. There is no guarantee that you will be able to factor a third degree polynomial.

One way is to look to see if it can be put into the form
(ax)^3 + (b)^3

This problem can. (3x)^3 + 2^3
All problems in this form have one factor of ax+b. So you just need to use synthetic division to solve the rest.

The factors are (3x + 2)(9x^2 - 6x + 4)
You put an extra x in front.

Answer 2

This is the sum of cubes.To factor, you must get the cubed root of 27x^3 which is 3x and of 8 which is 2. Then you substitute into the factoring formula and get :( 3x+2)(9x^2 -6x + 4) I think you have an extra x in your answer.

Answer 3

this is how I do it ....

step one.)

27x^3 , I just know that 3^3 = 27
also I just know that 2^3 = 8
thus I write

27x^3+8 = (3x)^3 + 2^3

step 2)

I forgot the question .... ah factorise...
I set y = 3x and write down

y^3 + 2^3 = 0

now I see that y = -2 is a solution of the above eq.

thus y+2 is a factor of y^3 + 2^3
now I try to find the other factor ....
(y+2)( ... ) = y^3 + 8

y + 2 | y^3 + 8 .............| y^2 -2y+4
...........y^3 +2y^2
...........---------------------
............8 -2y^2
............-4y-2y^2
...........-------------
.............8+4y
.............8+4y
............=========

So I find that
(y+2)( y^2 -2y+4 ) = y^3 + 8

now I try to factor the y^2 -2y+4
i dont see it , \I use the dumb abc formula and notice that this has no real factors.

So I am ready :
(y+2)( y^2 -2y+4 ) = y^3 + 8

putting back the y = 3x

gives the final factorisation
(3x+2)( 9x^2 -6x+4 ) = (3x)^3 + 8

Answer 4

ANS : (3x + 2)(9x^2 - 6x + 4)

Lets say you have

ax^3 + b

((ax^3)^(1/3) + b^(1/3))((ax^3)^(2/3) - ((ax^3)^(1/3) * b^(1/3)) + b^(2/3))

so you have 27x^3 + 8

((27x^3)^(1/3) + (8)^(1/3))((27x^3)^(2/3) - ((27x^3)^(1/3) * 8^(1/3)) + 8^(2/3)

(3x + 2)((3x)^2 - (3x * 2) + 2^2)

(3x + 2)(9x^2 - 6x + 4)

also unless your problem was x(27x^3 + 8), there's no x in front of x(3x + 2)(9x^2 - 6x + 4)

Answer 5

apply the (a+b)cube formula
SRRY ABT MY PREV ANSWER.. I DINT UNDERSTAND THE RAISED TO SIGN
the formula is
(a+b)^3= (a+b)(a^2-ab+b^2) ----here a =3x....ie...27 and b=2
=(3x+2)([3x]^2-3x*2+2^2)
=(3x+2)(9x^2-6x+4)

Answer 6

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

Put a = 3x and b = 2