How do you simplify #(3+x)/(2x^2+5x-3)#?

1 Answer
Aug 21, 2015

Answer:

#1/(2x-1)" "#, with #x!=1/2#

Explanation:

First, recognize that you're dealing with a rational expression, which means that you need to figure out what values of #x# will make the denominator equal to zero.

In your case, you need to find the values of #x# that will get you

#2x^2 + 5x - 3 = 0#

To do that, use the quadratic formula to find the two roots of this equation

#x_(1,2) = (-5 +- sqrt(5^2 - 4 * 2 * (-3)))/(2 * 2)#

#x_(1,2) = (-5 +- sqrt(49))/4#

#x_(1,2) = (-5 +- 7)/4 = {(x_1 = (-5 - 7)/4 = -3), (x_2 = (-5 + 7)/4 = 1/2) :}#

For a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

you can use the two roots to factor the equation

#color(blue)(ax^2 + bx + c = a * (x-x_1) * (x-x_2))#

In your case, you have #a=2# and can thus write

#2x^2 + 5x - 3 = 2 * (x-(-3)) * (x - 1/2)#

#" "= 2(x+3)(x-1/2)#

This means that your rational expression can be written as

#(color(red)(cancel(color(black)((x+3)))))/(2 * color(red)(cancel(color(black)((x+3)))) * (x-1/2)) = 1/(2 * (x-1/2))#

Notice that you still need to have #x!=1/2# in order to avoid having the denominator equal to zero.

Finally, rewrite the expression in a simpler form

#1/(2 * (x-1/2)) = 1/(color(red)(cancel(color(black)(2))) * ((2x-1))/color(red)(cancel(color(black)(2)))) = color(green)(1/(2x-1)," "x!=1/2)#