Question

simplify.
\\(\frac{w^{2}+4w-21}{36-4w^{2}}\\)

Explanation:

Step1: Factor the numerator and denominator

First, factor the quadratic in the numerator \(w^2 + 4w - 21\). We need two numbers that multiply to \(-21\) and add to \(4\). Those numbers are \(7\) and \(-3\). So, \(w^2 + 4w - 21=(w + 7)(w - 3)\).

Next, factor the denominator \(36 - 4w^2\). This is a difference of squares in the form \(a^2 - b^2=(a + b)(a - b)\), but first factor out the common factor of \(4\): \(36 - 4w^2 = 4(9 - w^2)=4(3 + w)(3 - w)\). Notice that \((3 - w)=-(w - 3)\), so we can rewrite the denominator as \(-4(w - 3)(w + 3)\) (or \(4(3 + w)(3 - w)\), but rewriting with \(-(w - 3)\) will help with canceling).

So now the expression is \(\frac{(w + 7)(w - 3)}{-4(w - 3)(w + 3)}\).

Step2: Cancel common factors

We can cancel the common factor of \((w - 3)\) from the numerator and the denominator (assuming \(w
eq3\) to avoid division by zero).

After canceling, we have \(\frac{w + 7}{-4(w + 3)}\), which can also be written as \(-\frac{w + 7}{4(w + 3)}\) or \(\frac{-w - 7}{4(w + 3)}\).

Answer:

\(-\dfrac{w + 7}{4(w + 3)}\) (or \(\dfrac{-w - 7}{4(w + 3)}\) or \(\dfrac{w + 7}{-4(w + 3)}\))