Simplify and State Restrictions (x^2-4x-5)/(x^2-3x-10) x (9x+3x^2/x^2-9)?

[(x^2 - 4x - 5)/(x^2 - 3x - 10)] × [(9x + 3x^2)/(x^2 - 9)]

= [(x^2 - 4x - 5)(9x + 3x^2)] / [(x^2 - 3x - 10)(x^2 - 9)]

= (3 (x + 1) x)/((x + 2) (x - 3)) (for x! = -3 and x! = 5)

[(x^2-4x-5)/(x^2-3x-10)]*(9x+3x^2/x^2-9)

=

[(x-5)(x+1)/(x-5)(x+2)]*(9x+3-9)

=

(x+1)(9x-3)/(x+2)

=

3(x+1)(3x-1)/(x+2).

Note that: (x^2-9) is not the same as x^2-9.

No. You need to properly use parentheses. The computer you will use in the future will insist upon proper terms.

As usual, taking you at your word:

(x² - 4x - 5)/(x² - 3x - 10) × (9x + 3x²/x² - 9)

= (x - 5)(x + 1)/[(x - 5)(x + 2)] × (9x + 3 - 9)

= (x + 1)/(x + 2) × (9x - 6)

= 3(3x - 2)(x + 1)/(x + 2), where x ≠ -2, x ≠ 0, x ≠ 5

f(x) = ((x^2 -4x - 5)/(x^2 -3x -10)) * ((9x + 3x^2)/(x^2-9))... assumed this is what you meant...

... x^2 -4x - 5 = (x -5)(x+1)

... x^2 -3x -10 = (x-5)(x+2) 

...... zero when x = 5 or x = -2

...... so can't have x = 5 or x = -2 (causes division by zero)

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

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

...... zero when x = -3 or x = 3

...... so can't have x = -3 or x = 3 (division by zero)

So, restrictions are x may not be -3, -2, 3 or 5

otherwise:

f(x)

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

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

= 3x(x+1)(3+x^2) / ((x+2)(x+3)(x-3)

So, you are wrong....