Question

How do you fully simplify `(x^2-8x+15)/(2x^2-7x-15)`?

Answer 1

`(x^2-8x+15)/(2x^2-7x-15) = (x-3)/(2x+3)`

with exclusion `x!=5`

Explanation:

Note that:

`x^2-8x+15 = x^2-8x+16-1`

`color(white)(x^2-8x+15) = (x-4)^2-1^2`

`color(white)(x^2-8x+15) = ((x-4)-1)((x-4)+1)`

`color(white)(x^2-8x+15) = (x-5)(x-3)`

So the numerator has zeros `x=3` and `x=5`

Are either of these zeros of the denominator?

`2(color(blue)(3)^2)-7(color(blue)(3))-15 = 18-21-15 = -18`

`2(color(blue)(5)^2)-7(color(blue)(5))-15 = 50-35-15 = 0`

So `x=5` is a zero and `(x-5)` a factor of the denominator:

`2x^2-7x-15 = (x-5)(2x+3)`

Hence:

`(x^2-8x+15)/(2x^2-7x-15) = ((color(red)(cancel(color(black)(x-5))))(x-3))/((color(red)(cancel(color(black)(x-5))))(2x+3))`

`color(white)((x^2-8x+15)/(2x^2-7x-15)) = (x-3)/(2x+3)`

with exclusion `x!=5`