Find a polynomial function ƒ(x) of degree 3 with real coefficients that satisfies the given conditions. Zeros of -3, 1, and 4; ƒ(2)=30

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Identify the general form of a cubic polynomial with the given zeros. Since the zeros are -3, 1, and 4, the polynomial can be written as $f(x) = a(x + 3)(x - 1)(x - 4)$, where $a$ is a real number coefficient to be determined.

Expand the factors partially or fully if needed to understand the polynomial structure, but keep the factorized form for now to make substitution easier.

Use the given condition $f(2) = 30$ to find the value of $a$. Substitute $x = 2$ into the polynomial: $f(2) = a(2 + 3)(2 - 1)(2 - 4)$.

Calculate the product inside the parentheses: $(2 + 3) = 5$, $(2 - 1) = 1$, and $(2 - 4) = -2$. So, $f(2) = a imes 5 imes 1 imes (-2) = -10a$.

Set the expression equal to 30 and solve for $a$: $-10a = 30$. Once $a$ is found, write the final polynomial function $f(x) = a(x + 3)(x - 1)(x - 4)$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polynomial Functions and Degree

A polynomial function is an expression involving variables raised to whole-number exponents with real coefficients. The degree of the polynomial is the highest exponent of the variable, which determines the general shape and number of roots. For a degree 3 polynomial, the function will have up to three roots and can be written in the form ƒ(x) = a(x - r1)(x - r2)(x - r3).

Zeros (Roots) of a Polynomial

Zeros of a polynomial are the values of x for which the function equals zero. They correspond to the x-intercepts of the graph. Given zeros allow us to express the polynomial as a product of factors (x - zero). For example, zeros at -3, 1, and 4 imply factors (x + 3), (x - 1), and (x - 4).

Using a Point to Find the Leading Coefficient

When the zeros are known, the polynomial can be written up to a constant multiplier a. To find a, substitute a given point (x, ƒ(x)) into the polynomial and solve for a. This ensures the polynomial passes through the specified point, such as ƒ(2) = 30, which helps determine the exact function.

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