To find a polynomial function $f(x)$ with zeros at $x = -3$, $x = 1$, and $x = 5$, and which also passes through the point $(7, 360)$, follow these steps:

Step 1: Set up the Polynomial Form

Since the zeros of the polynomial are given, we can express $f(x)$ as: $f(x) = a(x + 3)(x - 1)(x - 5)$ where $a$ is a constant that we'll determine using the given point $(7, 360)$.

Step 2: Substitute the Point $(7, 360)$

Since $f(7) = 360$, we substitute $x = 7$ and $f(x) = 360$ into the equation: $360 = a(7 + 3)(7 - 1)(7 - 5)$

Step 3: Simplify the Expression

Calculate each factor inside the parentheses: $360 = a(10)(6)(2)$ $360 = a \cdot 120$

Step 4: Solve for $a$

Divide both sides by 120 to isolate $a$: $a = \frac{360}{120} = 3$

Step 5: Write the Final Polynomial

Now that we know $a = 3$, we can write the polynomial as: $f(x) = 3(x + 3)(x - 1)(x - 5)$

Step 6: Expand the Polynomial (Optional)

Expanding this gives: $f(x) = 3(x^3 - 3x^2 - 13x + 15) = 3x^3 - 9x^2 - 39x + 45$

So, the polynomial function is: $f(x) = 3x^3 - 9x^2 - 39x + 45$

Would you like further details or have any questions?

Related Questions

1. How do we determine the degree of a polynomial based on its zeros?
2. What does the value of $a$ signify in the polynomial $f(x) = a(x + 3)(x - 1)(x - 5)$?
3. How would the polynomial change if it had a zero with multiplicity?
4. What are some methods to expand and simplify polynomial expressions?
5. How would we find the polynomial if it passed through a different point, say $(2, -40)$?

Tip

To verify your polynomial, substitute the zeros and the given point back into the polynomial to check consistency.