# Thread: Polynomials with Fractional Powers

1. ## Polynomials with Fractional Powers

Factor: x^(2/5) - 3x^(1/5) - 4 = 0

2. Originally Posted by magentarita
Factor: x^(2/5) - 3x^(1/5) - 4 = 0
Use a similar idea to that suggested here: http://www.mathhelpforum.com/math-he...lynomials.html

3. Originally Posted by magentarita
Factor: x^(2/5) - 3x^(1/5) - 4 = 0
Strictly speaking, that is not a polynomial: polynomials don't have fractional powers. But it can be made into one- let $y= x^{1/5}$ and solve for y first, then solve $x^{1/5}= y$ for x.

4. ## ok...

I know how to solve this equation using the substitution method but is there another way?

5. Originally Posted by magentarita
I know how to solve this equation using the substitution method but is there another way?
You have to realise that it's a quadratic.

Technically all methods for factorising involve this "substitution" - it's just whether you want to make it look a bit more pleasing to the eyes.

You could just as easily write it as

$(x^{\frac{1}{5}})^2 - 3x^{\frac{1}{5}} -4$

which becomes $(x^{\frac{1}{5}} - 4)(x^{\frac{1}{5}} + 1)$.

Originally Posted by Prove It
You have to realise that it's a quadratic.

Technically all methods for factorising involve this "substitution" - it's just whether you want to make it look a bit more pleasing to the eyes.

You could just as easily write it as

$(x^{\frac{1}{5}})^2 - 3x^{\frac{1}{5}} -4$

which becomes $(x^{\frac{1}{5}} - 4)(x^{\frac{1}{5}} + 1)$.

(5th root of x)^2 - 3(5th root of x) - 4 = 0

7. Originally Posted by magentarita
That's the exact same question, because $\sqrt[5]{x} = x^{\frac{1}{5}}$.
That's the exact same question, because $\sqrt[5]{x} = x^{\frac{1}{5}}$.