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

1 Answer

Answer:

#x^(-1/2)(x-1)^2=(x-1)^2/x^(1/2)=(x^(1/2)(x-1)^2)/x#

Explanation:

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

#ax^2+bx+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^(-1/2)# in the question, the difference is the #(-1/2)# 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^(-1/2)# term:

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

#x^(0-1/2)-2x^(1- 1/2)+x^(4/2-1/2)#

#x^0 xx x^(-1/2)-2x xx x^(-1/2)+x^2 xx x^(-1/2)#

#x^(-1/2)(1-2x+x^2)#

and now we can factor in the normal way:

#x^(-1/2)(x^2-2x+1)#

#x^(-1/2)(x-1)^2=(x-1)^2/x^(1/2)=(x^(1/2)(x-1)^2)/x#