Question

Last one?

Answer 1

since x + 1 = 0 is the root you are testing, then x = -1 needs to be your divisor

Answer 2

for part II) set up the coefficients in the synthetic division

Answer 3

three things to keep in mind:

1. keep the original signs (positive 12, positive 5, positive 3, negative 5)

Answer 4

2. the divisor goes in a box by itself on the left hand side

Answer 5

3. we have an x^4 term, and x^3 term, an x^2 term, but no x term so we must plug in a 0 as a placeholder between 3 and -5

Answer 6

with those three points in mind, try re-doing the setup

Answer 7

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
3. we have an x^4 term, and x^3 term, an x^2 term, but no x term so we must plug in a 0 as a placeholder between 3 and -5
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 8

good, now for part III carry out the division

Answer 9

good

then for part IV) is there a remainder?

Answer 10

no

Answer 11

so if there's a remainder, then is x + 1 a factor or nah?

Answer 12

um no

Answer 13

good, (x+1) is not a factor

and that's it

Answer 14

Thank you so much and Goodnight