MHB Factoring With Negative Powers

AI Thread Summary
The discussion focuses on factoring the expression (x^2 + 1)^(-2/3) + (x^2 + 1)^(-5/3). The correct approach involves factoring out the term with the smaller exponent, which is (x^2 + 1)^(-5/3). After factoring, the expression simplifies to (x^2 + 1)^(-5/3)((x^2 + 1) + 1). The final factored form is (x^2 + 1)^(-5/3)(x^2 + 2). The initial selection of the smallest power was noted as incorrect.
mathdad
Messages
1,280
Reaction score
0
Factor

(x^2 + 1)^(-2/3) + (x^2 + 1)^(-5/3)

Solution:

(x^2 + 1)^(-2/3)[1 + (x^2 + 1)^(2/5)]

Yes?
 
Mathematics news on Phys.org
We are given to factor:

$$\left(x^2+1\right)^{-\frac{2}{3}}+\left(x^2+1\right)^{-\frac{5}{3}}$$

So, we factor out the expression with the smaller exponent, observing that $$-\frac{5}{3}<-\frac{2}{3}$$...and then we subtract that exponent:

$$\left(x^2+1\right)^{-\frac{5}{3}}\left(\left(x^2+1\right)^{-\frac{2}{3}-\left(-\frac{5}{3}\right)}+\left(x^2+1\right)^{-\frac{5}{3}-\left(-\frac{5}{3}\right)}\right)$$

Now, simplify the exponents:

$$\left(x^2+1\right)^{-\frac{5}{3}}\left(\left(x^2+1\right)^{\frac{3}{3}}+\left(x^2+1\right)^{0}\right)$$

$$\left(x^2+1\right)^{-\frac{5}{3}}\left(\left(x^2+1\right)+1\right)$$

$$\left(x^2+1\right)^{-\frac{5}{3}}\left(x^2+2\right)$$
 
I selected the wrong smallest power.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Just chatting with my son about Maths and he casually mentioned that 0 would be the midpoint of the number line from -inf to +inf. I wondered whether it wouldn’t be more accurate to say there is no single midpoint. Couldn’t you make an argument that any real number is exactly halfway between -inf and +inf?

Similar threads

Replies
2
Views
1K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
5
Views
1K
Replies
3
Views
914
Replies
1
Views
1K
Back
Top