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.
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.
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)
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)
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...
To factor the expression 4x^3 - 36x + 16x^(-1), we first look for any common factors among the terms. In this case, all the terms have a factor of 4, so we can factor it out:
4(x^3 - 9x + 4x^(-1))
= 4(x^3 - 9x + 4/x)
Next, we need to look for possible rational roots using the Rational Root Theorem. The Rational Root Theorem states that any rational root of a polynomial equation will be of the form p/q, where p is a factor of the constant term (in this case, 4) and q is a factor of the leading coefficient (in this case, 1).
The possible factors of 4 are ±1, ±2, and ±4, and the possible factors of 1 are ±1. Therefore, the possible rational roots are: ±1, ±2, ±4.
To check which of these possible rational roots are actual roots of the equation, we can use synthetic division. Let's try checking if x = 1 is a root:
1 | 1 -9 4
_______
1 -8 -4
No remainder
Since there is no remainder, x = 1 is a root.
By performing synthetic division with x = 1, we get a quadratic equation:
(x - 1)(x^2 - 8x - 4)
Now, we can attempt to factor the quadratic equation. However, it doesn't factor nicely, so we can use the quadratic formula to find the remaining roots:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
For the quadratic equation x^2 - 8x - 4, a = 1, b = -8, and c = -4.
x = (-(-8) ± sqrt((-8)^2 - 4(1)(-4))) / (2(1))
x = (8 ± sqrt(64 + 16)) / 2
x = (8 ± sqrt(80)) / 2
x = (8 ± 4sqrt(5)) / 2
x = 4 ± 2sqrt(5)
Therefore, the factors of the expression 4x^3 - 36x + 16x^(-1) are:
4(x - 1)(x - (4 + 2sqrt(5)))(x - (4 - 2sqrt(5)))
Remember to consider the original exponents to obtain the final factored expression:
4(x^2 - x(sqrt(5)) - 1)(x - 1)