# Prove that 2x - 2/( x^2 + 3 ) = f(x) if f(x) = ( 2x^3 + 6x - 2 )/ ( x^2 + 3 )

### 2 Answers | Add Yours

The function f(x) = ( 2x^3 + 6x - 2 )/ ( x^2 + 3 )

f(x) = ( 2x^3 + 6x - 2 )/ ( x^2 + 3 )

2x - 2/( x^2 + 3 )

make the denominator the same for all the terms

[2x(x^2 + 3) - 2]/(x^2 + 3)

=> (2x^3 + 6x - 2)/ (x^2 + 3)

which is the same as f(x)

**This proves that 2x - 2/( x^2 + 3 ) = f(x)**

To prove that 2x - 2/( x^2 + 3 ) = f(x), we'll substitute f(x) by the given expression ( 2x^3 + 6x - 2 )/ ( x^2 + 3 ).We could manage the problem in 2 ways: either we'll calculate the left side, namely the difference 2x - 2/( x^2 + 3 ), or we'll re-write the expression of f(x).

We'll re-write f(x):

( 2x^3 + 6x - 2 )/ (x^2 + 3) = (2x^3 + 6x)/(x^2 + 3) - 2/(x^2 + 3)

We'll factorize by 2x the first ratio:

(2x^3 + 6x)/(x^2 + 3) = 2x(x^2 + 3)/(x^2 + 3)

We'll simplify the first ratio:

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

**f(x) = 2x - 2/(x^2 + 3)**

**We notice that the LHS = RHS, therefore 2x - 2/(x^2 + 3) = ****( 2x^3 + 6x - 2 )/ (x^2** **+ 3).**