Question

One factor of f(x) = x3 − 14x2 + 61x − 84 is (x − 7). What are the zeros of the function?

7, 4, −3
7, 4, 3
7, −4, −3
7, −4, 3

Answer 1

It may determine that there were no negative real roots through using Descartes' rule of the sign. Only one option with no negative noughts shown is the one mentioned below, and further calculation can be defined as follows:

Given:

`\bold{f(x) = x^3 - 14x^2 + 61x - 84}\\\\\bold{Factor=} (x-7)\\\\`

To find:

find zeros of function=?

Solution:

when factor=0

then

`\to \bold{x-7=0}\\\\\to \bold{x=7}\\\\`

putting the value of x into the given function:

`\to \bold{f(x) = x^3 - 14x^2 + 61x - 84}\\\\`

`\to \bold{f(7) = 7^3 - 14* 7^2 + 61* 7 - 84}\\\\\to \bold{f(7) = 343 - 14* 49 + 427 - 84}\\\\\to \bold{f(7) = 343 - 686 + 427 - 84}\\\\\to \bold{f(7) = 770- 770}\\\\\to \bold{f(7) =0}\\\\`

and When x=3:

`\to \bold{f(3) = 3^3 - 14* 3^2 + 61* 3 - 84}\\\\\to \bold{f(3) = 27 - 14* 9 + 183 - 84}\\\\\to \bold{f(3) = 27 - 126 + 183 - 84}\\\\\to \bold{f(3) = 210- 210}\\\\\to \bold{f(3) =0}\\\\`

and When x=4:

`\to \bold{f(4) = 4^3 - 14* 4^2 + 61* 4 - 84}\\\\\to \bold{f(4) = 64- 14* 16 + 244- 84}\\\\\to \bold{f(4) = 64 - 224 + 244- 84}\\\\\to \bold{f(3) = 308- 308}\\\\\to \bold{f(3) =0}\\\\`

Therefore the final choice "7,4,3".

• Please find garph in the attached file.

Answer 2

7, 4, 3

Explanation:

Using Descartes' rule of signs, you can determine there are zero negative real roots. The above answer choice is the only one with no negative zeros listed.

_____

The count of negative real roots is based on the number of sign changes when the signs of odd-degree terms are reversed. Doing that gives the signs of the terms as - - - -, so there are no changes. That means there are no negative real roots.

A graphing calculator confirms this choice.