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## anonymous one year ago Help? Which of the following is an actual zero of f(x) = x3 + x2 - 4x - 4? (4 points) 4 1 -1 -4

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1. jdoe0001

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

2. anonymous

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

3. jdoe0001

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

4. jdoe0001

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

5. jdoe0001

the one that gives 0, is the root

6. anonymous

@jdoe0001 i got 1

7. jdoe0001

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

8. jdoe0001

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

9. anonymous

so it wasnt one?

10. jdoe0001

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

11. anonymous

0

12. jdoe0001

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

13. anonymous

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

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