Question

table of contents > module 7: rational expressions and equations > module 7 pretest: rational expressio
module 7 pretest: rational expressions
equations
module 7 pretest: rational expressions and equations
score: 1/9 answered: 1/9
question 2
simplify \\(\frac{x^2 + 2x - 3}{x - 5} \times \frac{x - 3}{x - 1}\\)
hint: factor and reduce your fraction in order to fully simplify.

Explanation:

Step1: Factor the quadratic

Factor \(x^2 + 2x - 3\). We need two numbers that multiply to \(-3\) and add to \(2\). Those numbers are \(3\) and \(-1\), so \(x^2 + 2x - 3=(x + 3)(x - 1)\).
Now the expression becomes \(\frac{(x + 3)(x - 1)}{x - 5}\times\frac{x - 3}{x - 1}\).

Step2: Cancel common factors

Notice that \((x - 1)\) appears in the numerator of the first fraction and the denominator of the second fraction. We can cancel them (assuming \(x
eq1\) to avoid division by zero).
After canceling, we have \(\frac{(x + 3)}{x - 5}\times\frac{x - 3}{1}=\frac{(x + 3)(x - 3)}{x - 5}\).

Step3: Simplify the numerator

Using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\), where \(a = x\) and \(b = 3\), the numerator \((x + 3)(x - 3)=x^2 - 9\).
So the simplified expression is \(\frac{x^2 - 9}{x - 5}\) (or we can leave it as \((x + 3)(x - 3)\) over \(x - 5\), but expanding the numerator is also a valid simplification).

Answer:

\(\frac{x^2 - 9}{x - 5}\) (or \(\frac{(x + 3)(x - 3)}{x - 5}\))