solve for zeroes, y=(2x^2-2x-24)/x^2+3x-10

To find the zeroes of the function 

\[
y = \frac{2x^2 - 2x - 24}{x^2 + 3x - 10}
\]

we need to determine the values of \(x\) for which \(y = 0\). This happens when the numerator is equal to zero, provided that the denominator is not zero at those points.

### Step 1: Solve for the numerator to be zero

The numerator is:

\[
2x^2 - 2x - 24 = 0
\]

We can factor out a 2:

\[
2(x^2 - x - 12) = 0 \implies x^2 - x - 12 = 0
\]

Now, let's factor this quadratic expression:

\[
x^2 - x - 12 = (x - 4)(x + 3) = 0
\]

Setting each factor to zero gives:

\[
x - 4 = 0 \implies x = 4
\]
\[
x + 3 = 0 \implies x = -3
\]

So, the potential zeroes of the function are \(x = 4\) and \(x = -3\).

### Step 2: Check the denominator to ensure it is not zero

The denominator is:

\[
x^2 + 3x - 10 = 0
\]

To find where the denominator is zero, we solve this quadratic equation:

\[
x^2 + 3x - 10 = 0
\]

We can factor or use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = 3\), and \(c = -10\):

\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 40}}{2}
\]
\[
x = \frac{-3 \pm \sqrt{49}}{2}
\]
\[
x = \frac{-3 \pm 7}{2}
\]

This gives:

\[
x = \frac{-3 + 7}{2} = 2 \quad \text{and} \quad x = \frac{-3 - 7}{2} = -5
\]

So, the denominator is zero when \(x = 2\) and \(x = -5\).

### Step 3: Determine the valid zeroes of the function

The zeroes of the function occur when the numerator is zero, provided that the denominator is not zero at those points. From Step 1, the potential zeroes are \(x = 4\) and \(x = -3\).

Now, check if these values make the denominator zero:

- At \(x = 4\), the denominator is:

\[
4^2 + 3(4) - 10 = 16 + 12 - 10 = 18 \neq 0
\]

- At \(x = -3\), the denominator is:

\[
(-3)^2 + 3(-3) - 10 = 9 - 9 - 10 = -10 \neq 0
\]

Since the denominator is not zero at \(x = 4\) and \(x = -3\), both are valid zeroes of the function.

### Conclusion

The zeroes of the function \(y = \frac{2x^2 - 2x - 24}{x^2 + 3x - 10}\) are:

\[
\boxed{x = 4 \text{ and } x = -3}
\]

Would you like to explore more details or have any questions?

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Here are 5 related questions that might interest you:

1. How do you factor a quadratic equation?
2. What are the steps for solving a quadratic equation using the quadratic formula?
3. How can we determine if a function has any vertical asymptotes?
4. How do we find the domain and range of a rational function?
5. How can we simplify a rational function before solving for its zeroes?

**Tip:** Always check both the numerator and the denominator separately when dealing with rational functions to find their zeroes and any undefined points.

Learn how to solve for the zeroes of the rational function y = (2x^2 - 2x - 24)/(x^2 + 3x - 10). This problem involves quadratic equations, factoring, and the quadratic formula. Suitable for students in Grades 9-12.

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