# factorization of 128x4 - 54xy3please explain with the all steps

kjcdb8er | Certified Educator

Factorization means that you take what is common between different terms, and take that out of the parenthesis grouping those terms.

Factoring the following: 128x^4 - 54xy^3

First, note what is common between the two terms:

128 x^4 - 54 x y^3 =

2*7*11 x^4 - 3*3*3*2 x y^3

What is common between the two terms is one "2" and one "x". So take them out:

2x(7*11 x^3 - 3^3 y^3) =

2x(77x^3 - 27y^3)

neela | Student

Let A=128*x^4-54xy^3. We try to express A in as many possible factors as possible.

2x is a factor of both terms, 128x^4 and 54xy^3.Therefore, we can write A as :

A=2x(64x^3-27y^3)..............(1)

We know that 64x^3 = (4x)^3    and  27y^3 = (3y)^3.

Now let us have a transformation 4x=a and 3y=b.

Then 64x^3-27y^3 = a^3-b^3.

But a^3-b^3 = (a-b)(a^2+ab+b^2) is an identity.

64x^3-27y^3 = (4x)^3-(3y)^3= (4x-3y){(4x)^2+(4x)(3x)+(3y)^2}...................(2)

Replacing 64x^3  - 27 y^3   in  (1)  by  the expression on the right side of (2), we get:

A=2x(4x-3y)(16x^2-12xy+9y^2)

Therefore, 2x(4x-3y)(16x^2+12xy+9y^2) is the factor form of 128x^4-54xy^3

liona | Student

From the first post,

"First, note what is common between the two terms:

128 x^4 - 54 x y^3 =

2*7*11 x^4 - 3*3*3*2 x y^3

What is common between the two terms is one "2" and one "x". So take them out:

2x(7*11 x^3 - 3^3 y^3) =

2x(77x^3 - 27y^3)."

There is an error please. It should be 128x^4-54xy^3=2x(64x^3-27^3), in the first post.

That could further be factored also.