Question

Use synthetic division to find the quotient and the remainder for the problem: (12x^4 + 5^3 + 3x^2 - 5)/(x+1)?

Answer 1

Part I: What is the number you should use for the divisor in this division problem?
Part II: Set up the synthetic division problem by transferring the coefficients from the divided into a division table.
Part III: Carry out the division and write your answer as Quotient + (Remainder/Divisor).
Part IV: Is x+1 a factor of the polynomial (12x^4 + 5^3 + 3x^2 - 5)/(x+1)?

Answer 2

So when I did actually carried out the problem I got a remainder of +5

Answer 3

yes, that's correct. i just did it and go that too

Answer 4

Would (x+1) be the divisor and +5 as the remainder... Making Part III look like this?
\[(12x^4 + 5^3 + 3x^2 - 5)+\frac{ 5 }{ (x+1) }\]

Answer 5

That's right for part 3

Answer 6

wait noo... sorry.

Answer 7

So the equation already provided the divisor?

Answer 8

Okay, go ahead...

Answer 9

answer is 12 x^3 - 7x^2 + 10x - 10 + 5/(x+1)
use the bottom row of your synthetic division result

Answer 10

you got the last part right...+ 5/(x+1)

Answer 11

So this was my quotient... \[12 x^3 - 7x^2 + 10x - 10\] Is that what you mean?

Answer 12

yes.. that's the quotient they're talking about in Part 3

Answer 13

making the equation look like the instead? \[(12 x^3 - 7x^2 + 10x - 10) + \frac{ 5 }{ (x+1) }\]

Answer 14

That is precisely the answer to part 3...yes

Answer 15

At the end, it asks "Part IV: Is x+1 a factor of the polynomial (12x^4 + 5^3 + 3x^2 - 5)/(x+1)?" The answer is no, right?

Answer 16

correct....because you have to add that 5/(x+1) at the end. If you got 0 as the last value in your synthetic division instead of 5, it would be a factor.

Answer 17

Thank you! I was a bit unsure of myself

Answer 18

Sure! Good luck!