Week 13 – precalculus 11- Rational expression

This week in math I had a question on rational expressions that seemed very tricky at first. The question was:

\frac{3x}{x^2-64} + \frac{x-4}{8x + x^2}

then i start with the first step

1.factoring:

x^2 - 64 = (x-8)(x+8)

8x + x^2 = x(x+8)

The second step finding the Common denominator of them

x(x-8)(x+8)

and than write them with the common denominator all together

\frac{3x}{(x-8)(x+8)} = \frac{3x^2}{x(x-8)(x+8)}

\frac{x-4}{x(x+8)} = \frac{(x-4)(x-8)}{x(x-8)(x+8)}

after all these steps now we can do the adding or anything that the question asks:

\frac{3x^2 + (x-4)(x-8)}{x(x-8)(x+8)}

(x-4)(x-8) = x^2 - 12x + 32

3x^2 + x^2 - 12x + 32 = 4x^2 - 12x + 32

now we can see, we could still factor the numerator to 4

4(x^2 - 3x + 8)

And the final answer will be

\frac{4(x^2 - 3x + 8)}{x(x-8)(x+8)}

When working with rational expressions, we always need to check which numbers make the denominator zero.
Because division by zero is impossible → so those numbers are not allowed.

In this question:

\frac{3x}{x^2-64} + \frac{x-4}{8x + x^2}

1. First denominator:

x^2 - 64 = (x-8)(x+8)

If:

x = 8
or

x = -8

, the denominator becomes zero.

So these two values ​​are not allowed.

2. Second denominator:

8x + x^2 = x(x+8)

If:

x = 0
or

x = -8

, the denominator becomes zero.

(We already had x = -8)

3. Final result (Domain Restrictions)

Numbers that we cannot substitute for x:

x \neq 0
x \neq 8
x \neq -8

Because each of these values ​​causes the denominator to become zero.

\text{division by zero is undefined}

As soon as the denominator becomes zero, the whole expression becomes meaningless.

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