How do you "write the fraction as a sum of terms"?

Question

I'm stuck on this one section on my Algebra II/Trig summer assignment; the directions say "Write the fraction as a sum of terms." What does THAT mean? I tried searching the web for "sum of terms" but I get all these websites with confusing explanations. Can someone help me, please?

Example problem:

x^2 - 4x + 2 (fraction bar) x

^2 = second power/exponent 2

Answer 1

(x^2 - 4x + 2) / x can be rewritten as
x^2 / x - 4x / x + 2 / x
which is a "sum of terms"

Answer 2

This situation has lots of blind alleys, yet one thank you to show that's to contemplate a rational fraction of the style: a/(an-b) the place n > a > b are integers. Subtracting a million/n from it, we've: a/(an-b) - a million/n = (an - an + b)/(n(an-b)) = b/(an² - bn) the place we replace an² - bn with bn' - c, the place b > c. in this occasion, we've a the rest fraction the place b < a, and this technique would be repeated till we are left with a fragment of the style a million/n'', and the algorithmic technique ends with a finite sum of fractions. that's going to be stated that for any given rational fraction 0 < x < a million, there's a multiplicity of techniques that's expressed interior the style a million/a + a million/b + a million/c + ..., the place a, b, c... are all distinctive integers. This situation jogs my memory of Turing's Halting situation, in that what we are being asked to do is to show that if we subtract fractions of the style a million/n successively from the unique rational fraction, the technique will halt finally. that's style of exciting to objective doing it making use of random fractions, finally it halt with each and every so often some rather a great way out fractions. Addendum: If for any reason in the process the algorithmic technique we arise with 2 fractions that are alike, we are able to apply amir11elad's technique of changing between the fractions with 2 others, and if that motives yet another tournament, we are able to repeat, till after a finite style of step there'll be no extra effective suits and all the fractions would be distinctive. Addendum 2: right this is the info returned, extra concisely: a million) Subtract fractions of the style a million/n from the rational variety 0 < x < a million successively (with distinctive n) till we are left with a fragment of the style a/(an-b), the place n > a > b are integers. 2) Subtract a million/n from this fraction, so as that we are left with a fragment of the style b/(bn-c), the place n > b > c are integers. 3) proceed till we are left with the fraction of the style a million/n. in case you have a sequence of integers a > b > c > ...., finally, you will finally end up with a million. that's why the sequence will constantly be finite. Addendum 3: pay attention ye, Alexey V

Answer 3

You simply divide each term by the denominator, and then simplify:

(x^2 - 4x + 2)/x
= x^2/x - 4x/x + 2/x
= x - 4 + 2/x