# algebraic fraction

• September 17th 2009, 03:33 AM
mark
algebraic fraction
hi i've been told to do

$\frac {x^2 + 3x + 1}{x + 2}$ and the answer is $x + 1 - \frac{1}{x + 2}$

can someone show me how to arrive at this answer please? thanks
• September 17th 2009, 03:43 AM
Quote:

Originally Posted by mark
hi i've been told to do

$\frac {x^2 + 3x + 1}{x + 2}$ and the answer is $x + 1 - \frac{1}{x + 2}$

can someone show me how to arrive at this answer please? thanks

Hi

$
x^2+3x+1=(x+1)(x+2)-1
$

do you see how to go on from here ?
• September 17th 2009, 03:46 AM
Algebraic long division
Hello mark
Quote:

Originally Posted by mark
hi i've been told to do

$\frac {x^2 + 3x + 1}{x + 2}$ and the answer is $x + 1 - \frac{1}{x + 2}$

can someone show me how to arrive at this answer please? thanks

You need to use algebraic long division. There's an example here.

See if you can apply this to your example. You should get a quotient $x+1$ and a remainder $-1$. Can you then see what to do with the remainder? (Think about how you handle the remainder when you divide using ordinary numbers.

• September 17th 2009, 03:46 AM
mark
not really, thats how the book set it out before putting it in the form i had it in, could you show me from there please?
• September 17th 2009, 03:48 AM
mark
ok i'll try to figure out how to use the long division
• September 17th 2009, 04:04 AM
mark
hi could someone please write out the long division to answer this question so i can check my working?

thankyou
• September 17th 2009, 04:06 AM
mark
ah i figured out what math addict means now
• September 17th 2009, 04:28 AM
pacman
http://www.mathhelpforum.com/math-he...85d3376f-1.gif ?

without latex, ugly result

11111 __x_________
x + 2 / x^2 + 3x + 1
000000x^2 + 2x (subtract)
000000000000x + 1 ( bringdown 1)

then next term will be 1

00000 x + 1 0
x + 2 / x^2 + 3x + 1
000000x^2 + 2x (subtract)
000000000000x + 1 ( bringdown 1)
000000000000x + 2 (subtract)
00000000000000 - 1 ( remainder)

thus http://www.mathhelpforum.com/math-he...85d3376f-1.gif = http://www.mathhelpforum.com/math-he...642de82b-1.gif
• September 17th 2009, 04:38 AM
pacman

http://www.mathhelpforum.com/math-he...03b63781-1.gif, how?

x^2 + 3x + 1 = x^2 + 3x + (1 + 2 - 2), you breakdown the 3rd-term

then, rearrange (x^2 + 3x + 2) - 1, you will get

(x + 1)(x + 2) - 1

divide by (x + 2) each term

(x + 1)(x + 2)/(x + 2) - 1/(x + 2), notice the (x + 2) term cancels

thus, (x^2 + 3x + 1)/(x + 2) = (x + 1) - 1/(x + 2).

easy right?