Dividing polynomials...I have no idea what to do with this..

[erin524]

Please help!!!

2x/x^2-9 - 7/ x-3
______________________
3/x-3 + 2/x+3


Not sure if my typing is clear....it's fractions on top, with fractions on the bottom and a fraction line separating the top line from the bottom. Thank you.

 

[arthur ohlsten]

{2x/[x^2-9] - 7/[x-3]} / { 3/[x-3]-7/[x+3]}

let us simplify the top[numerator] first by placing over common denominator
===================================================
2x/[x^2-9] - 7[x-3] but x^2-9=[x-3][x+3]
2x/{[x-3][x+3]} - 7/[x-3]
[2x-7[x+3] ]/{[x-3][x+3]
[-5x-21] /{[x+3][x-3]}

================================================================
let us simplify the denominator
3/[x-3]+2/[x+3]
{3[x+3]+2[x-3]} /{[x-3][x+3]}
{3x+9+2x-6} / {[x-3][x+3]}
[5x+3]/{x-3][x+3]
===================================================================

to divide we invert the denominator and multiply
{[-5x-21] / {x+3][x-3] } times {[x-3][x+3] / [5x+3]
simplify
[-5x-21] / 5x+3] answer but a improper fraction. divde as you did in 5th grade

-1
______________
5x+3 l -5x-21
-5x-3
----------
-18

-1 - 18/[5x+3] answer

please check for math errors. The method is correct but the algebra might have a problem
a simple check might be to let x= a value [not 3 or -3] and see if you get the same answer from both the original and your final answer

let x=2
-1 -18/[5x+3] = -1 -18/13
[-13-18]/13
-31/13

2[2]/[4-9] -7/[-1] / {3/-1+2/5]
{4/[-5] +7} / {-3 + 2/5}
{[4-35]/-5 } /{ [-15+2]/5}
31/-13 or -31/13

checks at x=2 not sufficient but neccessary check

Arthur

 

[soroban]

Hello, erin524!

When given a complex fraction (one with more than two "levels"),
. . I prefer to eliminate the extra denominators immediately.

\(\displaystyle \text{Simplify: }\;\frac{\dfrac{2x}{x^2-9} - \dfrac{7}{x-3}} {\dfrac{3}{x-3} + \dfrac{2}{x+3}}\)

\(\displaystyle \text{[1] Find the LCD of the denominators.}\)

. . \(\displaystyle \text{For our problem, }\:LCD \:=\x-3)(x+3)\)


\(\displaystyle \text{[2] Multiply numerator and denominator by the LCD.}\)


\(\displaystyle \frac{(x-3)(x+3)\,\bigg[\dfrac{2x}{(x-3)(x+3)} - \dfrac{7}{x-3}\bigg]} {(x-3)(x+3)\,\bigg[\dfrac{3}{x-3} + \dfrac{2}{x+3}\bigg]}\)


. . \(\displaystyle = \;\frac{(\rlap{/////}x-3)(\rlap{/////}x+3)\!\cdot\!\dfrac{2x}{(\rlap{/////}x-3)(\rlap{/////}x+3)} \;-\; (\rlap{/////}x-3)(x+3)\!\cdot\!\dfrac{7}{\rlap{/////}x-3}} {(\rlap{/////}x-3)(x+3)\!\cdot\!\dfrac{3}{\rlap{/////}x-3} \;+\; (x-3)(\rlap{/////}x+3)\!\cdot\!\dfrac{2}{\rlap{/////}x+3}}\)


. . \(\displaystyle =\;\frac{2x-7(x+3)}{3(x+3) + 2(x-3)}\)


. . \(\displaystyle = \;\frac{2x - 7x - 21}{3x + 9 + 2x - 6}\)


. . \(\displaystyle = \;\frac{-5x-21}{5x+3}\)

 

[Denis]

{2x/[x^2-9] - 7/[x-3]} / { 3/[x-3]-7/[x+3]}

A "suggestion" (eliminates lotsa typing!):

x^2 - 9 = (x + 3)(x - 3)

let a = x+3 and b = x-3

[2x/(ab) - 7/b] / (3/b + 2/a)

Complete that, then substitute back in.

 

[erin524]

Thank you so much for all your help!! I love this website!!