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.
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.
4 answers
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)
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)
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)
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)
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...
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...