Question

How do you factor `x^(-3/2) + 2x^(-1/2) + x^(1/2)`?

Answer 1

`x^(-3/2)+2x^(-1/2)+x^(1/2) = x^(-3/2)(1+2x+x^2) = x^(-3/2)(x+1)^2`

Explanation:

`x^(a+b) = x^a*x^b`

`1+2x+x^2 = x^2 + 2x + 1 = (x+1)(x+1) = (x+1)^2`

Answer 2

`x^(-3/2) + 2 x^(-1/2) + x^(1/2) = ((x + 1) (x + 1))/x^(3/2)`

Explanation:

Starting with

`x^(color(brown)(-3/2)) + 2 x^(color(magenta)(-1/2)) + x^(color(orange)(1/2))`

extract the factor `x^(-3/2)` (using the rules for multiplication of indices---see below) to yield

`x^(color(red)(-3/2))(x^color(green)(0) + 2x^color(blue)(1) + x^color(purple)(2))`

Note that a common "forced" factor (here, `x^(color(red)(-3/2))`) has been taken outside of the brackets, to leave a "conventional" polynomial inside the brackets.

For the moment, to assist with the explanation, the terms in `x` in the polynomial have been left with explicit indices.

These will be tidied up in due course but before that, it might help to see what is happening in the "forced" extraction to compare the following set of additions:
`color(red)(-3/2) + color(green)(0) = color(brown)(-3/2)`
`color(red)(-3/2) + color(blue)(1) = color(magenta)(-1/2)`
`color(red)(-3/2) + color(purple)(2) = color(orange)(1/2)`

Noting that, in each case:

• the first addend is `color(red)(-3/2)`, which corresponds to the index of the term in `x` taken outside the bracket,
• the second addend (variously `color(green)(0), color(blue)(1) " and " color(purple)(2)`) corresponds to respective indices in the polynomial in the bracket,
• the third term (the sum) corresponds to the respective indices in the original expression.

Noting `x^0 = 1` and `x^1 = x`, this may be rewritten

`x^(-3/2)(1 + 2x + x^2) = (1+ 2x + x^2)/x^(3/2)`

Rearranging to make the polynomial more familiar

`= (x^2 + 2x + 1)/x^(3/2)`

From which you might see how to factorise the polynomial in the numerator as:

`((x + 1) (x + 1))/x^(3/2)`