# Calculate (x-2)/(x+1)+(x+3)/(x+2)-(5-x^2)/(x^2+3x+2).

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To calculate (x-2)/(x+1)+(x+3)/(x+2)-(5-x^2)/(x^2+3x+2).

We know the LCM of the denominators is (x+1)(x+2) = x^2+3x+2.

Thereforere let us write eachÂ term with equivalent term with LCM as denominator:

First term = (x-2)(x+2)/(x+1)(x+2) = (x^2 -4)/x+1)(x+2).

2nd term = (x+3)(x+1)/(x+1)(x+2) = (x^2+4x+3)/(x+1)(x+2).

3rd term = -(5-x^2)/(x+1)(x+2).

Now adding the three terms with their signs, we get:

Given expression = {x^2-4 +x^2+4x+3 -(5-x^2)}/(x+1)(x+2)

= {(x^2+x^2+x^2) +4x -4+3-5}/(x+1)(x+2)

= {3x^2+4x-6}/(x+1)(x+2)

Therefore (x-2)/(x+1)+(x+3)/(x+2)-(5-x^2)/(x^2+3x+2) = {3x^2+4x-6}/(x+1)(x+2)

To perform additions and subtractions of ratios, they must have the same denominator.

We'll have to calculate the least common denominator of the given ratios. For this reason, we'll use factorization for the denominator of the 3rd ratio.

We'll determine the roots of the expression x^2+3x+2:

x^2+3x+2 = 0

We'll apply the quadratic formula:

x1 = [-3 + sqrt(9 - 8)]/2

x1 = (-3+1)/2

x1 = -1

x2 = (-3-1)/2

x2 = -2

We'll re-write the expression x^2+3x+2 as a product of linear factors:

x^2+3x+2 = (x-x1)(x-x2)

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

We'll re-write the given expression:

E(x) = (x-2)/(x+1)+(x+3)/(x+2)-(5-x^2)/(x+1)(x+2)

We notice that the least common denominator of the 3 ratios is the denominator of the 3rd ratio:

LCD = (x+1)(x+2)

E(x) = [(x-2)(x+2)+(x+3)(x+1)-(5-x^2)]/(x+1)(x+2)

We'll remove the brackets. We'll recognize the difference of squares:

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

E(x) = [(x-2)(x+2)+(x+3)(x+1)-(5-x^2)]/(x+1)(x+2)

E(x) = (x^2 - 4 + x^2 + 4x + 3 - 5 + x^2)/(x+1)(x+2)

We'll combine like terms:

**E(x) = (3x^2 + 4x - 6)/(x+1)(x+2)**