Question

How do you factor `x^(7/3) - 3x^(4/3) + 2x^(1/3)`?

Answer 1

The Exp.`=x^(1/3)(x-1)(x-2)`.

Explanation:

Letting `x=y^3`, we find that,

The Expression`=(y^3)^(7/3)-3(y^3)^(4/3)+2(y^3)^(1/3)`

`=y^(3*7/3)-3y^(3*4/3)+2y^(3*1/3)`

`=y^7-3y^4+2y`

`=y(y^6-3y^3+2)`

`=y{(y^3)^2-3y^3+2}`

Observe that, the Poly. in `{...}` is a Quadratic in `y^3`, and,

The sum of its Co-effs.`=1-3+2=0`, meaning that, `y^3-1` is a factor

thereof. Therefore,

The Exp.`=y{ul((y^3)^2-y^3)-ul(2y^3+2)}`

`=y{y^3(y^3-1)-2(y^3-1)}`

`=y{(y^3-1)(y^3-2)}`

Since, `y^3=x, y=x^(1/3)`

Hence, the Exp.`=x^(1/3)(x-1)(x-2)`.

Enjoy Maths.!

Answer 2

`x^(1/3)(x-2)(x-3)`

Explanation:

In any factoring process, the first step is to find a common factor.

The fact that in each index there is a denominator of 3, is a clue that # x^(1/3) is a common factor which can be divided out of each term.

When the bases are the same and you are dividing, subtract the indices.

`x^(1/3)(x^(6/3) -3x^(3/3) + 2x^(0))`.

Simplifying leads to a simple quadratic trinomial.

`x^(1/3)(x^2 -3x+2)`

Read from right to left to get all the clues.

"Find factors of 2 which ADD to 3" `rarr 1xx2`
"The signs are the SAME, both negative"

`x^(1/3)(x-2)(x-3)`