Question

Help?
Which of the following is an actual zero of f(x) = x3 + x2 - 4x - 4? (4 points)

4

1

-1

-4

Answer 1

use the remainder theorem
if 4 is, then f(4) = 0
if 1 is, then f(1) = 0
if -1 is, then f(-1) =0
and if -4 is instead, then f(-4) =0

Answer 2

@jdoe0001 im so confused. It has -4 in the equation... would it be D?

Answer 3

-4 in the equation means nothing, in relation to a root
the remainder theorem only pertains to the constant in the binomial

Answer 4

so get the f(x) values, see which of the choices give you 0

Answer 5

the one that gives 0, is the root

Answer 6

@jdoe0001 i got 1

Answer 7

so... what's 1 again? or...what's f(1) ?

Answer 8

\(\bf f(x)=x^3+x^2-4x-4
\\ \quad \\
f({\color{brown}{ 1}})=({\color{brown}{ 1}})^3+({\color{brown}{ 1}})^2-4({\color{brown}{ 1}})-4\implies f(1)\ne 0\)

Answer 9

so it wasnt one?

Answer 10

well. what did you get for f(1)? assume that you did it

Answer 11

0

Answer 12

\(\bf f(x)=x^3+x^2-4x-4
\\ \quad \\
f({\color{brown}{ 1}})=({\color{brown}{ 1}})^3+({\color{brown}{ 1}})^2-4({\color{brown}{ 1}})-4\implies f(1)=1+1-4-4
\\ \quad \\
f(1)=2-8\implies f(1)=-6\)

Answer 13

but i guess i wrote it out wrong, you had it a different way