A degree 3 polynomial f(x) has zeros at -3, 10, and 6, and the coefficient of x³ equals 5. Find an equation for f(x).

Step-by-step solution

1. 1

   Turn each zero into a factor. The Factor Theorem says $f(r) = 0$ exactly when $(x - r)$ divides $f(x)$. Zeros $-3$, $10$ and $6$ therefore contribute the factors $(x+3)$, $(x-10)$ and $(x-6)$. Watch the sign: the zero $-3$ gives $\bigl(x-(-3)\bigr) = (x+3)$.

2. 2

   Account for the unknown scale factor. Three linear factors already give degree $3$, but they are not the whole story: any constant multiple has the same zeros. So the most general form is

   $f(x) = a(x+3)(x-10)(x-6)$

3. 3

   Pin down $a$ using the leading coefficient. Multiplying out, the only way to reach $x^3$ is to take $x$ from each factor, so the $x^3$ term is $a \cdot x \cdot x \cdot x = ax^3$. Matching the given coefficient of $x^3$:

   $a = 5 \quad\Longrightarrow\quad f(x) = 5(x+3)(x-10)(x-6)$

4. 4

   Expand to standard form. First multiply the two factors that pair off cleanly:

   $(x+3)(x-10) = x^2 - 7x - 30$

   Then multiply by the remaining factor:

   $(x^2 - 7x - 30)(x-6) = x^3 - 13x^2 + 12x + 180$

   Finally distribute the $5$:

   $f(x) = 5x^3 - 65x^2 + 60x + 900$

5. 5

   Check that the three zeros really vanish. Substituting each root into the standard form gives $f(-3) = f(6) = f(10) = 0$, and the coefficient of $x^3$ is $5$ as required. Both the factored and expanded versions are correct answers — the factored form shows the zeros, the standard form shows the coefficients.