Here is the content with "fract"s removed:
1) Domain Restrictions for the Following Polynomials: a) For the expression \( \frac{x}{x-4} \), the restriction comes from the denominator, which cannot be zero. Therefore, we set \( x - 4 \neq 0 \), resulting in the restriction \( x \neq 4 \).
b) In the expression \( \frac{x-8}{x^2(x+3)} \), we have two factors in the denominator that cannot equal zero. First, \( x^2 \neq 0 \), which gives us \( x \neq 0 \). Second, \( x + 3 \neq 0 \) leads to \( x \neq -3 \). Thus, the overall restrictions are \( x \neq 0 \) and \( x \neq -3 \).
c) For the expression \( 5 - x \), the restriction is that \( 5 - x \neq 0 \), which simplifies to \( x \neq 5 \).
d) The expression \( x^2 - 5x + 6 \) can be factored into \( (x-2)(x-3) \). The restrictions come from setting the factors not equal to zero, leading us to \( x - 2 \neq 0 \) and \( x - 3 \neq 0 \). Therefore, the restrictions are \( x \neq 2 \) and \( x \neq 3 \).
2) Operations with Rational Expressions - Simplification:
- For the expression \( \frac{5}{x+3} + \frac{2}{x-2} \), we need a common denominator, which is \( (x + 3)(x - 2) \):
\[ \frac{5(x - 2) + 2(x + 3)}{(x + 3)(x - 2)} = \frac{5x - 10 + 2x + 6}{(x + 3)(x - 2)} = \frac{7x - 4}{(x + 3)(x - 2)} \]
- For \( \frac{x - 5}{x^2 - 3x - 10} \), first we factor the denominator:
\[ x^2 - 3x - 10 = (x - 5)(x + 2) \]
Now we simplify:
\[ \frac{x - 5}{(x - 5)(x + 2)} = \frac{1}{x + 2} \quad (x \neq 5) \]
- For the expression \( \frac{x^2 - 4}{x^2 + 4x - 12} \), we can factor both the numerator and denominator:
\[ x^2 - 4 = (x - 2)(x + 2) \] \[ x^2 + 4x - 12 = (x + 6)(x - 2) \]
Thus, we have:
\[ \frac{(x - 2)(x + 2)}{(x + 6)(x - 2)} = \frac{x + 2}{x + 6} \quad (x \neq 2) \]
- For \( \frac{3x^2}{3x - 6x} \), we simplify the denominator:
\[ 3x - 6x = -3x \quad (x \neq 0) \]
So, we have:
\[ \frac{3x^2}{-3x} = -x \]
- To solve the equation \( \frac{x}{5} + \frac{x^2 + 2x - 8}{4} = 1 \), we find a common denominator. The least common multiple of 5 and 4 is 20. Multiplying all terms by 20 gives:
\[ 20\left(\frac{x}{5}\right) + 20\left(\frac{x^{2} + 2x - 8}{4}\right) = 20(1) \]
This simplifies to:
\[ 4x + 5(x^{2} + 2x - 8) = 20 \]
Distributing gives:
\[ 4x + 5x^{2} + 10x - 40 = 20 \]
Collecting like terms results in:
\[ 5x^{2} + 14x - 40 - 20 = 0 \]
which simplifies to:
\[ 5x^{2} + 14x - 60 = 0 \]
We apply the quadratic formula with \( a=5 \), \( b=14 \), and \( c=-60 \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 5 \cdot (-60)}}{2 \cdot 5} \]
The discriminant calculates as:
\[ 14^2 - 4 \cdot 5 \cdot (-60) = 196 + 1200 = 1396 \]
Thus, we have:
\[ x = \frac{-14 \pm \sqrt{1396}}{10} = \frac{-14 \pm 2\sqrt{349}}{10} = \frac{-7 \pm \sqrt{349}}{5} \]
The solutions are:
\[ x = \frac{-7+\sqrt{349}}{5} \quad \text{or} \quad x = \frac{-7-\sqrt{349}}{5} \]