# Write the simplified fraction [(x+2)^2*(x+1)/(x+1)^2]*[(x^2+2x+1)/(x^2+3x+2)]

### 2 Answers

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to simplify: [(x+2)^2*(x+1)/(x+1)^2]*[(x^2+2x+1)/(x^2+3x+2)]

[(x+2)^2*(x+1)/(x+1)^2]*[(x^2+2x+1)/(x^2+3x+2)]

=> [(x+2)^2*(x+1)*(x^2+2x+1)/(x+1)^2*(x^2+3x+2)]

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

=> [(x+2)^2*(x+1)*(x + 1)^2/(x+1)^2*(x + 1)(x + 2)]

cancel common terms in the numerator and denominator

=> (x+2)

The fraction simplifies to x + 2

Top Answer

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll write the numerators and denominators as distinct factors. For this reason, we'll determine the roots of the 2nd numerator:

x^2 + 2x + 1 = 0

We'll recognize the perfect square:

(x+1)^2 = 0

Now, we'll determine the roots of the denominator of the 2nd fraction:

﻿﻿x^2 + 3x + 2 = 0

We'll apply quadratic formula:

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

x1 = (-3+1)/2

x1 = -1

x2 = (-3-1)/2

x2 = -2

The equation will be written as:

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

We'll re-write the factorised expression:

[(x+2)^2*(x+1)/(x+1)^2]*[(x+1)^2/(x + 1)(x + 2)]

We'll cancel common factors:

[(x+2)^2*(x+1)/(x+1)^2]*[(x+1)^2/(x + 1)(x + 2)] = x + 2

The  requested simplified result of the given expression is: (x+2)