Can someone simplify this improper rational function 2x^2-4x/x+1?

Question

My book used long division to convert this problem into the sum of a polynomial and a rational function but I need to know the steps they used to get there.

This is what they gave me:

2x^2-4x/x+1 simplified to 2x-6 + 6/x+1

Answer 1

I donot know in which standard you are studying,but I would request you to follow any of the following processes to solve these odd problems
2x^-4x divided by x+1
=2x(x+1)-6(x+1)+6
So the quotient is 2x-6 and the remainder is 6.
You can write the answer of the division as
2x-6 +6/(x+1)

another process is the application of the remainder theorem.It states that if any polynomial is divided by(x+a) and if x+a=0 i.e x= -a,then if you substitute x by -a in the polynomial,the value of the polynomial is the remainder of actual division
You are dividing 2x^2-4x by x+1.Let x+1=0 or x=-1
therefore,2x^2-4x
=2*(-1)^2-4(-1)
=2+4=6
Hence 6 will be the remainderif you divided
2x^2-4x by x+1
Now subtract 6 from 2x^-4x and you get2x^2-4x-6 which will be exactly divisible by x+1
2x^2-4x-6/(x+1)
=2(x^2-2x-3)/(x+1)
=2(x+1)(x-3)/(x+1)
Hence,the quotient is 2(x-3) and the remainder as earlier calculated is 6 and this can be written as 2x-6+6/(x+1)

All said and done ,the best way to solve the problem is in the following way
2x^2-4x/x+1
=(2x^2-4x-6+6)/(x+1)
={(2x-6)(x+1)+6}/(x+1)
=2x-6 +(6/x+1)

Answer 2

(2x^2-4x)/(x+1)
Divide x into 2x^2 and get 2x as the 1st term in your answer.
Now multiply the 2x by (x+1) getting 2x^2 +2x.
Subtract ttnis from 2x^2 -4x getting -6x
Now divide -6x by x getting -6 as the 2nd term in your answer.
Now multiply -6 by(x+1) getting -6x -6.
Now subtract -6x-6 from -6x getting 6. This is your remainder.
So your answer is 2x-6 +6/(x+1).

Note that you put the remainder over the divisor (x+1).

It's just like saying 5/3 = 1 with remainder 2 = 1 2/3

Answer 3

1st question: = (2x² + 2x - 4)/(x² - 5x - 14) = (2[x² + x - 2])/([x - 7][x + 2]) = (2[{x + 2}{x - 1}])/([x - 7][x + 2]) = (2[x - 1])/(x - 7) Answer: (2[x - 1])/(x - 7) 2nd question: = ([x² - 3x - 10]/[x² - 2x - 15])(x² + 10x + 21) = ([{x - 5}{x + 2}]/[{x - 5}{x + 3}])([x + 7][x + 3]) = ([x + 2]/[x + 3])([x + 7][x + 3]) = (x + 2)(x + 7) Answer: (x + 2)(x + 7) or x² + 9x + 14

Answer 4

(2x^2-4x)/(x+1)
= [2(x+1)^2 - 8x - 2] / (x+1)
= 2(x+1) - [8x+2]/(x+1)
=2x+2 - {[8(x+1)-6]/(x+1)}
=2x+2 - {8 - 6/(x+1)}
=2x+2 - 8 + 6/(x+1)
=2x-6 +6/(x+1)

Answer 5

(2x^2 - 4x) / (x+1)
= (2x(x+1) -2x -4x) / (x+1)
= 2x - (6x / (x+1) )
= 2x - ( (6x +6 - 6 ) / (x+1) )
= 2x - ( (6(x+1) - 6) / (x+1) )
= 2x - 6(x+1)/(x+1) + (6 / (x+1))
= 2x -6 + 6/(x+1)

Answer 6

2x^2 - 4x / (x+1)
[ 2x^2 (x+1) - 4x ] / (x+1)
[ 2x^3 + 2x^2 - 4x ] / (x+1)
2x [ x2 + x - 2 ] / (x+1)
2x ( x + 2 ) ( x - 1 ) / (x+1)

Answer 7

2x^2-4x/x+1
{2x^2(x+1)-4x}/(x+1)
{2x^3+2x^2-4x}/(x+1)
2x(x^2+x-2)/(x+1)
2x{x^2-x+2x-2}/(x+1)
2x{x(x-1)+2(x-1)}/(x+1)
2x(x-1)(x+2)/(x+1) ans

Answer 8

2x^2-4x
x+1(2x)
-----------
2x^2 - 4x
2x^2 + 2x
-------------
-6x
x+1(-6)
-------------
-6x
-6x - 6
--------------
-6


2x - 6 - 6/(x+1)

Answer 9

2x(x-2)/x+1


That's what I would say =\

Answer 10

no