Question: (\[ \frac{x^2 + 6x + 9}{x^2 + 2x - 3} \]\)…

\(\[ \frac{x^2 + 6x + 9}{x^2 + 2x - 3} \]\)

Solution

To simplify the expression \[ \frac{x^2 + 6x + 9}{x^2 + 2x - 3} \] we need to factor both the numerator and the denominator. Factor the numerator \(x^2 + 6x + 9\). The numerator is a perfect square trinomial: \[ x^2 + 6x + 9 = (x + 3)^2 \] Factor the denominator \(x^2 + 2x - 3\). To factor the quadratic, find two numbers that multiply to \(-3\) and add to \(2\). The numbers are \(3\) and \(-1\). So, \[ x^2 + 2x - 3 = (x + 3)(x - 1) \] Now rewrite the original expression using the factors: \[ \frac{(x + 3)^2}{(x + 3)(x - 1)} \] Cancel the common factor \((x + 3)\) from the numerator and the denominator: \[ \frac{x + 3}{x - 1} \] The simplified expression is \[ \frac{x + 3}{x - 1} \]