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

##### 1 Answer

${x}^{- \frac{1}{2}} {\left(x - 1\right)}^{2} = {\left(x - 1\right)}^{2} / {x}^{\frac{1}{2}} = \frac{{x}^{\frac{1}{2}} {\left(x - 1\right)}^{2}}{x}$

#### Explanation:

Normally, when dealing with a trinomial, we're looking for something that looks like:

$a {x}^{2} + b x + c$

and our question doesn't have that! What it does have, however, if we compare the $c$ term in what we are normally looking for to the ${x}^{- \frac{1}{2}}$ in the question, the difference is the $\left(- \frac{1}{2}\right)$ exponent. And, in fact, all the terms in the question have that same difference between what we expect over what we have. So let's first factor out a ${x}^{- \frac{1}{2}}$ term:

${x}^{- \frac{1}{2}} - 2 {x}^{\frac{1}{2}} + {x}^{\frac{3}{2}}$

${x}^{0 - \frac{1}{2}} - 2 {x}^{1 - \frac{1}{2}} + {x}^{\frac{4}{2} - \frac{1}{2}}$

${x}^{0} \times {x}^{- \frac{1}{2}} - 2 x \times {x}^{- \frac{1}{2}} + {x}^{2} \times {x}^{- \frac{1}{2}}$

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

and now we can factor in the normal way:

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

${x}^{- \frac{1}{2}} {\left(x - 1\right)}^{2} = {\left(x - 1\right)}^{2} / {x}^{\frac{1}{2}} = \frac{{x}^{\frac{1}{2}} {\left(x - 1\right)}^{2}}{x}$