Question

precalculus - unit 1 review name ______ section b - rational expressions & equations 3. simplify (\frac{x^2 - 16}{x^2 - 9} cdot \frac{x^2 + x - 6}{x^2 - 4})

Explanation:

Step1: Factor each quadratic expression

We use the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\) and factoring trinomials.

• For \(x^2-16\), we have \(x^2 - 16=(x + 4)(x - 4)\) (since \(16 = 4^2\))
• For \(x^2-9\), we have \(x^2 - 9=(x + 3)(x - 3)\) (since \(9=3^2\))
• For \(x^2+x - 6\), we need two numbers that multiply to \(- 6\) and add to \(1\). The numbers are \(3\) and \(-2\), so \(x^2+x - 6=(x + 3)(x - 2)\)
• For \(x^2-4\), we have \(x^2 - 4=(x + 2)(x - 2)\) (since \(4 = 2^2\))

So the original expression \(\frac{x^2-16}{x^2-9}\cdot\frac{x^2+x - 6}{x^2-4}\) becomes \(\frac{(x + 4)(x - 4)}{(x + 3)(x - 3)}\cdot\frac{(x + 3)(x - 2)}{(x + 2)(x - 2)}\)

Step2: Cancel out common factors

We can cancel out the common factors in the numerator and the denominator.

• The \((x + 3)\) in the denominator of the first fraction and the numerator of the second fraction cancels out.
• The \((x - 2)\) in the numerator of the second fraction and the denominator of the second fraction cancels out.

After canceling, we are left with \(\frac{(x + 4)(x - 4)}{(x - 3)(x + 2)}\)

We can expand the numerator: \((x + 4)(x - 4)=x^2-16\)

So the simplified form is \(\frac{x^2-16}{(x - 3)(x + 2)}\) or we can expand the denominator: \((x - 3)(x + 2)=x^2 - x-6\), so it can also be written as \(\frac{x^2-16}{x^2 - x - 6}\)

Answer:

\(\frac{(x + 4)(x - 4)}{(x - 3)(x + 2)}\) (or \(\frac{x^2-16}{x^2 - x - 6}\))