# Thread: Long & synthetic division

1. ## Long & synthetic division

Hi Guys!

I'm new - very glad I found this. I have a prescribed textbook for my math course - in the back they only give you answers to the problems but do not show how they got the answer - so its difficult to trace your steps and see where you went wrong. Which explains me being stuck with these 3 problems.

I'll explain :

Long or synthetic division can be used for this one:

2x^4 - X^3 + 9x^2 divided by x^2 + 4

When I do the above one, I get to a point where I have powers that arent the same so I can't minus - all the ways I try it I either get stuck where I can't divide more or I get a wrong answer than that in the book - I'd like to see what answer some of you get, maybe the book is wrong. Please type our your steps - Im desperate to see where I go wrong.

Okay this one you have to use long division - again I get a horribly wrong answer if compared to the back of the book:

6x^3 + 2x^2 + 22x divided by 2x^2 + 5

The one below must be done using synthetic division - again my answer is wrong compared to the back of my textbook and I can't seem to understand why or how:

x^3 + 2x^2 + 2x + 1 divided by x + 2

I would appreciate your help a lot I'm either overlooking something or the book is wrong and Im very keen to find out which!

Thanks!

Janine

2. Hello, Janine!

You need to leave spaces for "missing terms" . . .

$\displaystyle (2x^4 - x^3 + 9x^2) \div (x^2 + 4)$

. . $\displaystyle \begin{array}{ccccccc} & & & & 2x^2 & -x & +1 \\ & & -- & -- & -- & -- & -- \\ x^2+4 & ) & 2x^4 & -x^3 & +9x^2 \\ & & 2x^4 & & + 8x^2 \\ & & -- & -- & -- \\ & & & -x^3 & +x^2 \\ & & & -x^3 & & -4x \\ & & & -- & -- & -- \\ & & & & x^2 & + 4x \\ & & & & x^2 & & +4 \\ & & & & -- & -- & -- \end{array}$
. . . . . . . . . . . . . . . . . . . . $\displaystyle 4x\quad-4$

Answer: . $\displaystyle 2x^2 - x + 1 + \frac{4x-4}{x^2+4}$

$\displaystyle (6x^3 + 2x^2 + 22x) \div (2x^2 + 5)$

. . $\displaystyle \begin{array}{cccccc} & & & & 3x &+ 1 \\ & & -- & -- & -- & -- \\ 2x^2+5 & ) & 6x^3 & +2x^2 &+ 22x \\ & & 6x^3 & & +15x \\ & & -- & -- & -- \\ & & & 2x^2 & +7x \\ & & & 2x^2 & & +5 \\ & & & -- & -- & -- \\ & & & & 7x & -5 \end{array}$

Answer: . $\displaystyle 3x + 1 + \frac{7x-5}{2x^2 + 5}$

$\displaystyle (x^3 + 2x^2 + 2x + 1) \div (x + 2)$

. . $\displaystyle \begin{array}{cccccc} -2 & | & 1 & 2 & 2 & 1 \\ & | & & \text{-}2 & 0 & \text{-}4 \\ & & --&--&--&-- \\ & & 1 & 0 & 2 & \text{-}3 \end{array}$

Answer: . $\displaystyle x^2 + 2 - \frac{3}{x+2}$