Precalculus

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 4, Problem 56RE

To determine

To calculate:

All the real zeroes of the following functions of polynomial with the help of Rational zeroes theorem and the factor $f$ over the real zeroes:

$f(x)=4x3−4x2−7x−2$

Expert Solution & Answer

Answer to Problem 56RE

The real zeroes of the given polynomial function are $f(x)=4x3−4x2−7x−2$ are $2,−12$

Explanation of Solution

Given information:

$f(x)=4x3−4x2−7x−2$

Formula Used:

The potential rational zeroes is given by the following expression:

$x=±pq$

Where,

Quotients of the factors of the last term: $p$

The factors of the leading coefficient: $q$

Calculation:

The given function is:

$f(x)=4x3−4x2−7x−2$

The degree of the given polynomial function is $3$ which shows that $f$ must have at most $3$ real zeroes.

Now, the values of $p,q$ are defined as:

$p=2q=4$

Factors of $p$ are $±1,±2$ and factors of $q$ is $±1,±2,±4$ .

So the potential rational zeroes are defined as:

$pq=±1,±2,±12,±14$

Now test the potential rational zeroes with the help of the synthetic division:

If the value of remainder is zero $f(1)=0$ then $1$ is zero of $f$ and $x−1$ is a factor.

$14−4−7−240−740−7−9$

Here, the remainder is $f(1)=−9$ . That’s why $1$ is not zero of $q1$ .

Now test for $−1$ by using the synthetic division:

$−14−4−7−2−4−81548−1513$

Here, the remainder is $f(−1)=13$ . That’s why $−1$ is not zero of $q1$ .

Now test for $2$ by using the synthetic division:

$24−4−7−28824410$

Here, the remainder is $f(2)=0$ . That’s why $2$ is zero of $q1$ and $x−2$ is a factor. Use the bottom row of synthetic division to factor $f$ :

$f(x)=4x3−4x2−7x−2f(x)=(x−2)(4x2+4x+1)$

The solution of the given polynomial equation $f(x)=4x3−4x2−7x−2$ are the zeroes of the $q1(x)=4x2+4x+1$ .

The potential rational zeroes of $q1(x)=4x2+4x+1$ :

$pq=±1,±12,±14$

Now test the potential rational zeroes with the help of the synthetic division:

Now test for $−1$ :

$144148489$

Here, the remainder is $f(1)=9$ . That’s why $1$ is not zero of $q1$ .

Now test for $−1$ by using the synthetic division:

$−1441−40401$

Here, the remainder is $f(−1)=1$ . That’s why $−1$ is not zero of $q1$ .

Now test for $12$ by using the synthetic division:

$1244123464$

Here, the remainder is $f(12)=4$ . That’s why $12$ is not zero of $q1$ .

Now test for $−12$ by using the synthetic division:

$−12441−2−1420$

Here, the remainder is $f(12)=0$ . That’s why $12$ is zero of $q1$ and $x+12$ is a factor. Use the bottom row of synthetic division to factor $f$ :

$f(x)=4x3+4x2−7x+2f(x)=(x−2)(4x2−4x+1)f(x)=(x−2)(x+12)(4x−2)f(x)=4(x−2)(x+12)2$

The real zeroes of $f$ are $−2,12$ . When factored, $f$ is

$f(x)=4x3+4x2−7x+2f(x)=4(x−2)(x+12)2$

Hence, real zeroes are $2,−12$ .

Chapter 4, Problem 55RE

Chapter 4, Problem 57RE

Chapter 4 Solutions

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All the real zeroes of the following functions of polynomial with the help of Rational zeroes theorem and the factor f over the real zeroes: f ( x ) = 4 x 3 − 4 x 2 − 7 x − 2