Question

for exercises 47–54, use the remainder theorem to determine if the given number c is a zero of the polynomial. (see example 7 on p. 218)

1. ( f(x) = x^4 + 3x^3 - 7x^2 + 13x - 10 )

a. ( c = 2 ) b. ( c = -5 )

1. ( g(x) = 2x^4 + 13x^3 - 10x^2 - 19x + 14 )

a. ( c = -2 ) b. ( c = -7 )

Explanation:

Problem 47a:

Step1: Recall Remainder Theorem

The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( x - c \), the remainder is \( f(c) \). So, to check if \( c = 2 \) is a zero of \( f(x)=x^{4}+3x^{3}-7x^{2}+13x - 10 \), we need to find \( f(2) \).

Step2: Substitute \( x = 2 \) into \( f(x) \)

$$
LATEXBLOCK0
$$

Wait, that's not zero. Wait, maybe I made a mistake. Wait, let's recalculate:

$$
LATEXBLOCK1
$$

Wait, maybe the problem is 47a, but let's check again. Wait, maybe I miscalculated. Wait, \( 2^4=16 \), \( 3\times2^3 = 3\times8 = 24 \), \( -7\times2^2=-7\times4=-28 \), \( 13\times2 = 26 \), \( -10 \). So \( 16 + 24=40 \), \( 40-28 = 12 \), \( 12 + 26=38 \), \( 38-10 = 28 \). So \( f(2)=28
eq0 \)? Wait, maybe the polynomial is different? Wait, the polynomial is \( x^4 + 3x^3-7x^2 + 13x - 10 \). Wait, maybe I made a mistake. Wait, let's try \( c = 1 \): \( 1 + 3-7 + 13-10=0 \). Oh, maybe the problem is 47a with \( c = 2 \) is wrong? Wait, no, the user provided the problem. Wait, maybe I misread the polynomial. Let me check again: \( f(x)=x^4 + 3x^3-7x^2 + 13x - 10 \).

Wait, maybe the original problem has a typo, but according to the remainder theorem, we calculate \( f(c) \). If \( f(c)=0 \), then \( c \) is a zero.

Problem 47b:

Step1: Recall Remainder Theorem

We need to check if \( c=-5 \) is a zero of \( f(x)=x^{4}+3x^{3}-7x^{2}+13x - 10 \), so we find \( f(-5) \).

Step2: Substitute \( x=-5 \) into \( f(x) \)

$$
LATEXBLOCK0
$$

So \( f(-5)=0 \), so \( c = -5 \) is a zero.

Problem 48a:

Step1: Recall Remainder Theorem

We need to check if \( c=-2 \) is a zero of \( g(x)=2x^{4}+13x^{3}-10x^{2}-19x + 14 \), so we find \( g(-2) \).

Step2: Substitute \( x=-2 \) into \( g(x) \)

$$
LATEXBLOCK0
$$

Wait, that's not zero. Wait, maybe I made a mistake. Let's recalculate:

\( (-2)^4 = 16 \), \( 2\times16 = 32 \); \( (-2)^3=-8 \), \( 13\times(-8)=-104 \); \( (-2)^2 = 4 \), \( -10\times4=-40 \); \( -19\times(-2)=38 \); \( +14 \). So \( 32-104=-72 \), \( -72-40=-112 \), \( -112 + 38=-74 \), \( -74 + 14=-60 \). So \( g(-2)=-60
eq0 \)? Wait, maybe the polynomial is different. Wait, the polynomial is \( 2x^4 + 13x^3-10x^2-19x + 14 \). Let's try \( c = 1 \): \( 2 + 13-10-19 + 14=0 \). So \( c = 1 \) is a zero.

Problem 48b:

Answer:

\( c = 2 \) is not a zero (since \( f(2)=28
eq0 \))

47b: