Question

Please help with this factoring problem:

2x^3/2 − 6x^1/2 + 4x^−1/2

what i did first was make all the exponents whole numbers, so I got:

4x^3-36x+16x^-1

I don't know what to do from there.

Answer 1

You can't just make the exponents whole numbers the way you do for a polynomial.

You can't just square each term individually. After all,

(x+y)^2 is not the same as x^2 + y^2
If you plug in, say, x=9 and evaluate the original expression vs your proposed alteration, you will see they are not at all related.

What you can do is make a substitution, say, u = x^1/2, and then you have

2u^3 - 6u + 4/u

2/u (u^4 - 3u^2 + 2)
2/u (u^2-1)(u^2-2)
2/u (u-1)(u+1)(u-√2)(u+√2)
Now, since u = √x, we have

2/√x (√x-1)(√x+1)(√x-√2)(√x+√2)

An unusual kind of problem.

Answer 2

Okay, you cannot just square all the terms. You need to factor out a common multiple. In this case x^(−1/2).

The remaining term will be a quadratic. You can leave it like that, or better, use the quadratic root formular to factorise. Then simplify.

2x^(3/2) − 6x^(1/2) + 4x^(−1/2)

= 2 x^(-1/2)×x^(2) - 6x^(-1/2)×x + 4x^(-1/2)

= x^(-1/2) (2x^2 - 6 x +1)

= x^(-1/2) (x-(3-√7)/2)(x-(3+√7)/2)

= (2x-3+√7)(2x-3-√7)/(4√x)

Answer 3

Edit: Made a typo in one line.

2x^(3/2) − 6x^(1/2) + 4x^(−1/2)

= 2 x^(-1/2)×x^(2) - 6x^(-1/2)×x + 4x^(-1/2)

= x^(-1/2) (2x^2 - 6 x + 4)

= x^(-1/2) 2(x-2)(x-1)

= 2(x-2)(x-1)/(4√x)

Answer 4

Good one, Graham. How silly of me not to notice that

2/u (u^2-1)(u^2-2) = 2/√x (x-1)(x-2)

duh. Still, we disagree by a factor of 1/4, but I'm sure Maria can resolve that...