The table below shows several values of x and the corresponding values of the third-degree polynomial function f. | x | f(x) | |---|---| | -1 | 0 | | 2 | 0 | | 5 | 0 | | 0 | 30 | If f(x) = a(x-r_1)(x-r_2)(x-r_3) for all real numbers x, where a, r_1, r_2, and r_3 are constants, what is the value of a

The table below shows several values of x and the corresponding values of the third-degree… — SAT

The table below shows several values of $x$ and the corresponding values of the third-degree polynomial function $f$ .

If $f(x) = a(x-r_1)(x-r_2)(x-r_3)$ for all real numbers $x$ , where $a$ , $r_1$ , $r_2$ , and $r_3$ are constants, what is the value of $a$ ?

Answer: 3

Since the function

f(x)

is a third-degree polynomial with roots at

x = -1

x = 2

, and

x = 5

, it can be factored as

f(x) = a(x+1)(x-2)(x-5)

. Evaluating this expression at

x = 0

gives

f(0) = a(1)(-2)(-5) = 10a

. From the table,

f(0) = 30

, so setting

10a = 30

yields

a = 3

Identify the roots and factors of the polynomial

f(x)

using the table.

The roots are

x = -1

x = 2

, and

x = 5

, which correspond to the factors

(x+1)

(x-2)

, and

(x-5)

Points where

f(x) = 0

represent the

x

-intercepts (roots) of the function.

Write the general form of the cubic polynomial using its factors.

f(x) = a(x+1)(x-2)(x-5)

A third-degree polynomial with three distinct real roots can be factored completely as

a(x-r_1)(x-r_2)(x-r_3)

Substitute the point

(0, 30)

into the equation to solve for the constant coefficient

a

a = 3

The table gives

f(0) = 30

, which allows us to set up the equation

30 = a(0+1)(0-2)(0-5)

and solve for

a

Using the relationship between the factors, roots, and points on the graph of a polynomial function to determine its equation.