Glencoe Algebra 2 Student Edition C2014

1st Edition

ISBN: 9780076639908

Author: McGraw-Hill Glencoe

Publisher: MCG

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Question

Chapter SH, Problem 5.20EP

To determine

Find the zeros of the given function.

Expert Solution & Answer

Answer to Problem 5.20EP

The function has zeros at $−1,1,−2i$ and $2i$ .

Explanation of Solution

Given:

The given function is $f(x)=x4+3x2−4$ .

Calculation:

Since $f(x)$ has degree $4$ , the function has $4$ zeros.

Find the type of zeros.

Examine the number of sign changes for $f(x)$ and $f(−x)$ .

$f(x)=x4+3x2−4 no yes$

Because there are one sign changes for the coefficients of $f(x)$ , the function has one or zero positive real zeros.

$f(−x)=(−x)4+3(−x)2−4=x4+3x2−4 no yes$

Because there is one sign changes for the coefficients of $f(−x)$ , the function has one or zero negative real zeros.

Thus, $f(x)$ has two real zerosand two imaginary zeros or four imaginary zeros.

Determine the real zeros. List some possible values, and then use synthetic substitution to evaluate $f(x)$ for real values of $x$ .

$x$	$1$	$0$	$3$	$0$	$−4$
$−2$	$1$	$−2$	$7$	$−14$	$24$
$−1$	$1$	$−1$	$4$	$−4$	$0$
$0$	$1$	$0$	$3$	$0$	$−4$
$1$	$1$	$1$	$4$	$4$	$0$
$2$	$1$	$2$	$7$	$14$	$24$

From the table, zero occurs at $x=−1$ and $x=1$ . Since there are one negative real zero and one positive real zeros.

Use synthetic substitution with depressed polynomial $f(x)=x3+x2+4x+4$ to find two imaginary zeros.

Substitute $f(x)=0$ .

$x3+x2+4x+4=0$

Use factorization method.

$x3+x2+4x+4=0⇒x2(x+1)+4(x+1)=0⇒(x+1)(x2+4)=0⇒x2+4=0⇒x2=−4⇒x=±−4i2=−1⇒x=±22i2⇒x=±2i$

Hence the function haszeros at $−1,1,−2i$ and $2i$ .

Chapter SH, Problem 5.19EP

Chapter SH, Problem 5.21EP

Chapter SH Solutions

Glencoe Algebra 2 Student Edition C2014

Ch. SH - Prob. 1.1EP Ch. SH - Prob. 1.2EP Ch. SH - Prob. 1.3EP Ch. SH - Prob. 1.4EP Ch. SH - Prob. 1.5EP Ch. SH - Prob. 1.6EP Ch. SH - Prob. 1.7EP Ch. SH - Prob. 1.8EP Ch. SH - Prob. 1.9EP Ch. SH - Prob. 1.10EP

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