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

Question

2016 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-19T14:43:45

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

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 · 2016-08-26T08:16:13

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

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)#