Question

This is going to require a bit of time ;)
Synthetically divide: f(x) = 2x^3 + 4x^2 + 2x - 3 – 4 divided by x + 2

Answer 1

not much time really, its just adding and multiplying

Answer 2

I just need it set up :/ I know you put 2 in the box, what else?

Answer 3

3 2 1 0
2x^3 + 4x^2 + 2x - 3 – 4; this end is wierd, is it a typo?

Answer 4

and my setup tends to be different than a textbook ..... a personal preference type thing

Answer 5

Nope, it has to have a remainder of 4. SHould I change it to just -4 instaead of -3-4?

Answer 6

well, -3-4 = -7 ... but are you saying the division needs to end in a -4?

Answer 7

Yes, it has to have a remainder of 4

Answer 8

Sorry, a remainder of -4

Answer 9

first thing i do is make sure all the x parts are there .... high to low in sequential order

3 2 1 0 ; they seem to be accounted for
2x^3 + 4x^2 + 2x - 3 – 4


I dont remove the x parts, i use them as a visual cue, since they are redundant anyhow. Since we are dividing by x+2, then when x=-2 we get a zero, so lets divide this by the -2

2x^3 + 4x^2 + 2x - 3 – 4
0 <-- add this row
--------------------------
-2 | <-- multiply this row



2x^3 + 4x^2 + 2x - 3 – 4
0 -4 0 <-- add this row
--------------------------
-2 | 2 0 2 <-- multiply this row


like that, ill let you work out the last parts

Answer 10

Wow. I so got it now! That's not even sarcasm, I'll write this down as a reference later. Thanks for your help *medal*

Answer 11

good luck :)

Answer 12

just as a wrap up, the numbers in the bottom row are the coeffs of the "qoutient" polynomial; which is where my visual cue of the x parts helps me out.

2x^3 + 4x^2 + 2x - 3 – 4
0 -4 0
--------------------------
-2 | 2 0 2 (......)
x^2 x c R

\[\frac{2x^3 + 4x^2 + 2x - 3 – 4}{x+2}~=~ 2x^2 + 2 + \frac{R}{x+2}\]