Question: Find all zeros of the function: $t(x) = x^2 + 2x - 48$. The zeros are $x = $
1. **State the problem:** We need to find all zeros of the function $$t(x) = x^2 + 2x - 48$$ which means solving for $x$ when $t(x) = 0$.
2. **Set the function equal to zero:**
$$x^2 + 2x - 48 = 0$$
3. **Use the quadratic formula or factorization.** Here, we try to factor:
We look for two numbers that multiply to $-48$ and add to $2$.
4. **Find factors:**
The numbers $8$ and $-6$ satisfy this because $8 \times (-6) = -48$ and $8 + (-6) = 2$.
5. **Write the factored form:**
$$x^2 + 2x - 48 = (x + 8)(x - 6) = 0$$
6. **Solve each factor for zero:**
$$x + 8 = 0 \Rightarrow x = -8$$
$$x - 6 = 0 \Rightarrow x = 6$$
7. **Conclusion:** The zeros of the function are $$x = -8$$ and $$x = 6$$.
These are the points where the graph of $t(x)$ crosses the x-axis.