Simplify

Problem

(2*x2−3*x−2)/(x2−1)/(2*x2+5*x+2)/(x2+x−2)

Solution

1. Rewrite the division of two fractions as the multiplication of the first fraction by the reciprocal of the second fraction.

(2*x2−3*x−2)/(x2−1)⋅(x2+x−2)/(2*x2+5*x+2)

2. Factor the quadratic trinomial 2*x2−3*x−2 into (2*x+1)*(x−2)

2*x2−3*x−2=(2*x+1)*(x−2)

3. Factor the difference of squares x2−1 into (x−1)*(x+1)

x2−1=(x−1)*(x+1)

4. Factor the quadratic trinomial x2+x−2 into (x+2)*(x−1)

x2+x−2=(x+2)*(x−1)

5. Factor the quadratic trinomial 2*x2+5*x+2 into (2*x+1)*(x+2)

2*x2+5*x+2=(2*x+1)*(x+2)

6. Substitute the factored forms back into the expression.

((2*x+1)*(x−2))/((x−1)*(x+1))⋅((x+2)*(x−1))/((2*x+1)*(x+2))

7. Cancel the common factors (2*x+1) (x−1) and (x+2) from the numerator and the denominator.

(x−2)/(x+1)

Final Answer

(2*x2−3*x−2)/(x2−1)/(2*x2+5*x+2)/(x2+x−2)=(x−2)/(x+1)

Want more problems? Check here!