Question

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

Answer 1

`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)`