How do you divide: (2x^3 + 8x^2 - 14x - 20) ÷ (x+5) ?1?

Answer 1

i just did the problem and i found it easier to do long division,

x + 5 (imagine division thing here) 2x^3 + 8x^2 - 14x - 20

now what you do first is see what multiplied by x can give you 2x^3, which is 2x^2 then you multiply that 2x^2 by 5

and just do that until you cancel out everything this problem works out to where theres no remainder

heres the work i scanned in, hope it helps =]


http://i75.photobucket.com/albums/i291/acarpy12/mathscan.jpg

Answer 2

You do it the same way as you do long division.
. . . . 2x^2
x + 5)2x^3 + 8x^2 - 14x - 20

. . . . 2x^2
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . . 2x^3 + 10x^2

. . . . 2x^2
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . .-2x^3 - 10x^2
. . . . . . . .. - 2x^2 - 14x - 20

. . . . 2x^2. - 2x
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . .-2x^3 - 10x^2
. . . . . . . .. - 2x^2 - 14x - 20

. . . . 2x^2. - 2x
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . .-2x^3 - 10x^2
. . . . . . . .. - 2x^2 - 14x - 20
. . . . . . . . - 2x^2 - 10x

. . . . 2x^2. - 2x
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . .-2x^3 - 10x^2
. . . . . . . .. - 2x^2. - 14x - 20
. . . . . . . . + 2x^2 + 10x
. . . . . . . . . . . . . . . - 4x - 20

. . . . 2x^2. - 2x. . . - 4
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . .-2x^3 - 10x^2
. . . . . . . .. - 2x^2. - 14x - 20
. . . . . . . . + 2x^2 + 10x
. . . . . . . . . . . . . . . - 4x - 20

. . . . 2x^2. - 2x. . . - 4
x + 5)2x^3 +. 8x^2 - 14x - 20
. . . .-2x^3 - 10x^2
. . . . . . . .. - 2x^2. - 14x - 20
. . . . . . . . + 2x^2 + 10x
. . . . . . . . . . . . . . . - 4x - 20
. . . . . . . . . . . . . . . - 4x - 20

With enough practice you can go with "short division", not actually writing down all the operations:

. . . . 2x^2. - 2x. . . - 4
x + 5)2x^3 +. 8x^2 - 14x - 20

"Synthetic division" leaves out the x's:

. . . . 2. - 2 . - 4
1 + 5)2 +. 8 - 14 - 20
. . . .-2 - 10
. . . . . . .- 2. - 14 - 20
. . . . . . + 2 + 10
. . . . . .. . . . . - 4 - 20
. . . . . .. . . . . - 4 - 20