Question

the solutions of $x^2 - 2x - 24 = 0$ are:
8 and -3
-8 and 3
-6 and 4
6 and -4
question 30
1 pts
simplify: $\frac{7x - 21}{x^2 + 7x + 10} cdot \frac{x^2 - 25}{x - 3}$
$\frac{7(x + 5)}{x + 2}$
$\frac{(7x - 21)(x - 5)}{(x + 2)(x - 3)}$
$\frac{7(x - 5)}{x + 2}$
$-\frac{35}{2}$

Explanation:

First Problem (Quadratic Equation)

Step1: Factor the quadratic

We find two numbers that multiply to $-24$ and add to $-2$. These numbers are $-6$ and $4$.
$x^2 -2x -24 = (x-6)(x+4)=0$

Step2: Solve for $x$

Set each factor equal to 0.
$x-6=0 \implies x=6$; $x+4=0 \implies x=-4$

Step1: Factor all terms

Factor numerators and denominators:
$7x-21=7(x-3)$; $x^2+7x+10=(x+2)(x+5)$; $x^2-25=(x+5)(x-5)$
Expression becomes: $\frac{7(x-3)}{(x+2)(x+5)} \cdot \frac{(x+5)(x-5)}{x-3}$

Step2: Cancel common factors

Cancel $(x-3)$ and $(x+5)$ from numerator/denominator.
$\frac{7\cancel{(x-3)}}{(x+2)\cancel{(x+5)}} \cdot \frac{\cancel{(x+5)}(x-5)}{\cancel{x-3}} = \frac{7(x-5)}{x+2}$

Answer:

6 and -4

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Second Problem (Rational Expression Simplification)