Question

How do you simplify `(x^2 – 3x – 10)/(x^2 – 9)div(x^2 + 4x + 4)/(x+3)`#?

Answer 1

First, factor your quadratic functions. Finding its roots will lead us to the factors and then it'll be way easier to calculate your result. So, let's find the roots (using Bhaskara)!

`f(x)=x^2-3x-10`

`(3+-sqrt(9-4(1)(-10)))/2`
`(3+-7)/2`
`x_1=5`, the same as the factor `x-5=0`
`x_2=-2`, the same as the factor `x+2=0`

`f(x)=x^2-3x-10=(x-5)(x+2)`

Next: `g(x)=x^2-9`

`x^2=9`
`x=+-3`, which generate the factors `(x-3)(x+3)`

Finally, `h(x)=x^2+4x+4`, which will have one root: `x=-2` and then give us the same factor twice: `(x+2)(x+2)`

Rewriting:

`(((x-5)(x+2))/((x-3)(x+3)))/(((x+2)(x+2))/(x+3))`=`((x-5)cancel((x+2))cancel((x+3)))/((x-3)cancel((x+3))cancel((x+2))(x+2))`=`(x-5)/((x-3)(x+2))`=`(x-5)/(x^2-x-6)`

Answer 2

This explanation's going to be all math. I'm assuming you know how to factor polynomials, but if you don't, then don't be afraid to comment to ask how; I don't bite! Okay, let's start.

`(x^2–3x–10)/(x^2–9) ÷ (x^2+4x+4)/(x+3)`

We can flip the 2nd fraction and multiply it.

`(x^2–3x–10)/(x^2–9) times (x+3)/(x^2+4x+4)`

Now let's factor the trinomials into binomials. Hopefully this will let us cancel out some stuff.

`((x-5)(x+2))/(x^2-9) times (x+3)/((x+2)(x+2))`

We can also factor `x^2 - 9` into `x+3` and `x-3`.

`((x-5)(x+2))/((x-3)(x+3)) times (x+3)/((x+2)(x+2))`

Now, I can cancel out some of the factors to simplify the answer.

`(x-5)/(x-3) times 1/(x+2) = (x-5)/((x-3)(x+2))=(x-5)/(x^2 - x -6)`

This should be the final answer, unless your teacher prefers to show the factors of trinomials, in which case `(x-5)/((x-3)(x+2))` is also acceptable.