Question

1. - / 1 points simplify the complex fraction. \\(\dfrac{\left(\dfrac{x^2 + 10x + 24}{4x^2 - 20x + 25}\

ight)}{\left(\dfrac{x^2 - 3x - 28}{4x^2 - 25}\
ight)}\\) \\(, x \
eq -4, x \
eq -\dfrac{5}{2}\\) resources read it tutorial

Explanation:

Step1: Factor all quadratics and binomials

Factor \(x^2 + 10x + 24\): \(x^2 + 10x + 24=(x + 4)(x + 6)\)
Factor \(4x^2 - 20x + 25\): \(4x^2 - 20x + 25=(2x - 5)^2\)
Factor \(x^2 - 3x - 28\): \(x^2 - 3x - 28=(x - 7)(x + 4)\)
Factor \(4x^2 - 25\): \(4x^2 - 25=(2x - 5)(2x + 5)\)

The complex fraction becomes:
\(\frac{\frac{(x + 4)(x + 6)}{(2x - 5)^2}}{\frac{(x - 7)(x + 4)}{(2x - 5)(2x + 5)}}\)

Step2: Divide by multiplying by reciprocal

Dividing by a fraction is multiplying by its reciprocal:
\(\frac{(x + 4)(x + 6)}{(2x - 5)^2}\times\frac{(2x - 5)(2x + 5)}{(x - 7)(x + 4)}\)

Step3: Cancel common factors

Cancel \((x + 4)\) from numerator and denominator:
Cancel one \((2x - 5)\) from \((2x - 5)^2\) and \((2x - 5)\):

We get: \(\frac{(x + 6)(2x + 5)}{(2x - 5)(x - 7)}\)

Answer:

\(\frac{(x + 6)(2x + 5)}{(2x - 5)(x - 7)}\)