Question: User: f(x) = x³ + ax² - ax + 48, where a is a constant
Given that f(-6) = 0, find the value of a.
(2 marks)
a =
1. **State the problem:** We are given the function $$f(x) = x^3 + ax^2 - ax + 48$$ where $$a$$ is a constant. We know that $$f(-6) = 0$$ and need to find the value of $$a$$.
2. **Use the given condition:** Substitute $$x = -6$$ into the function and set it equal to zero:
$$f(-6) = (-6)^3 + a(-6)^2 - a(-6) + 48 = 0$$
3. **Simplify the expression:**
$$-216 + 36a + 6a + 48 = 0$$
Combine like terms:
$$-216 + 42a + 48 = 0$$
4. **Simplify constants:**
$$-168 + 42a = 0$$
5. **Solve for $$a$$:**
Add 168 to both sides:
$$42a = 168$$
Divide both sides by 42 (showing cancellation):
$$\frac{\cancel{42}a}{\cancel{42}} = \frac{168}{42}$$
So,
$$a = 4$$
**Final answer:** $$a = 4$$