# What is the maximum integer value of the fraction ( 2x^2+4x+5 )/( x^2+2x+2 ) if x is real number ?

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### 1 Answer

We'll note the fraction as E(x).

E(x) = (2x^2+4x+5)/(x^2+2x+2)

We'll re-write the numerator in this way:

2x^2+4x+5 = x^2 + 2x + 2 + x^2 + 2x + 2 + 1

Now, we'll group the terms from the right side such as to create trinomials:

2x^2+4x+5 = (x^2+2x + 2) + (x^2+2x+2) + 1

We'll divide both sides by (x^2+2x+2):

(2x^2+4x+5)/(x^2+2x+2) = (x^2+2x + 2)/(x^2+2x+2) + (x^2+2x+2)/(x^2+2x+2) + 1/(x^2+2x+2)

We'll simplify and we'll get:

(2x^2+4x+5)/(x^2+2x+2) = 1 + 1 + 1/(x^2+2x+2)

The result of the addition of the terms from the right side is an integer number, if and only if the fraction 1/(x^2+2x+2) is an integer number, also.

For 1/(x^2+2x+2) to be integer, (x^2+2x+2) = 1

Shifting 1 to the left, we'll get:

x^2 + 2x + 1 = 0

We notice that we've get a perfect square:

(x+1)^2 = 0

x = -1

The expression will become:

E(x) = 1 + 1 - 1

E(x) = 1

**The maximum integer value of the given fraction, if x is a real number, is E(x) = 1.**