practice7 sec 3.6, 4.1, 4.2

1]
The function is 1-1. Find

SOLUTION

The domain and the range of the function are

.
Set

and solve for

To eliminate the cube-root, raise both sides of the equation to the third power.

Now, following the process of finding

, we replace the

on the left side of the equation by

and we replace the

on the side of the equation by

The domain and the range are

.

2]
Using synthetic division and the Remainder Theorem ,
find

given that

.

SOLUTION
By synthetic division,

The third line gives the result of the synthetic division process.
According to the Remainder Theorem, the last number of the third line, the remainder, is the
value of the given polynomial evaluated at

. That is,

.

3]
Find all the roots exactly (rational, irrational, and imaginary) of the polynomial equation

.

SOLUTION
By the rational Roots Theorem, the possible rational roots are

.
By synthetic division,

is a root of the equation and, by the Factor Theorem,

The second factor is quadratic and we know how to solve quadratic equations.

By the quadratic formula,

.

The roots of the third degree polynomial equation are

.

4]
     Use synthetic division table and the
               Location Theorem:
                     If

is continuous on an interval

and

are two numbers
in

, and

and

are of opposite sign, then there is at

least one

intercept between

and

.

Locate each real zero of the polynomial

between successive integers

SOLUTION
The following table gives the results of the synthetic division for the integers 2, 3, 4, 5, 6, 7.

The last number in each line (after the first line)
is the value of the polynomial evaluated at the given point.

changes sign in the intervals

, and

.

We have located three real zeros as being within each of the above intervals.
Since the polynomial is of the third degree, it has only three zeros ,
so we have located all of them.

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© edmond 2003


		Algebra B Practice 7 Module 3 & 4 Problems relating to sec 3.6, 4.1, 4.2

	1] The function is 1-1. Find . SOLUTION The domain and the range of the function are . Set and solve for . To eliminate the cube-root, raise both sides of the equation to the third power. Now, following the process of finding , we replace the on the left side of the equation by and we replace the on the side of the equation by . The domain and the range are . 2] Using synthetic division and the Remainder Theorem , find given that . SOLUTION By synthetic division, The third line gives the result of the synthetic division process. According to the Remainder Theorem, the last number of the third line, the remainder, is the value of the given polynomial evaluated at . That is, . 3] Find all the roots exactly (rational, irrational, and imaginary) of the polynomial equation . SOLUTION By the rational Roots Theorem, the possible rational roots are , . By synthetic division, So is a root of the equation and, by the Factor Theorem, The second factor is quadratic and we know how to solve quadratic equations. By the quadratic formula, . The roots of the third degree polynomial equation are . 4] Use synthetic division table and the Location Theorem: If is continuous on an interval , and are two numbers in , and and are of opposite sign, then there is at least one intercept between and . Locate each real zero of the polynomial between successive integers SOLUTION The following table gives the results of the synthetic division for the integers 2, 3, 4, 5, 6, 7. The last number in each line (after the first line) is the value of the polynomial evaluated at the given point. changes sign in the intervals , , and . We have located three real zeros as being within each of the above intervals. Since the polynomial is of the third degree, it has only three zeros , so we have located all of them. top © edmond 2003