Can anyone perform this operation and simplify (x^2-2x-15 / x^2 + x –6) / (x^2 + x –6 / x^2 –2x –35?

Question

Can anyone perform this operation and simplify
X^2 – 2x –15 x^2 + x –6
---------------- * ----------------
x^2 + x –6 x^2 –2x –35

Answer 1

(x^2-2x-15 / x^2 + x –6) / (x^2 + x –6 / x^2 –2x –35?= [(x-5)(x+2)/(x+3)(x-2)]/[(x+3)(x-2)/(x-7)(x+5)]
= (x-5)(x+3)/(x-7)(x+5) (because (x+3)(x-2)'s canceled out)

Answer 2

a million. bear in mind that with absolute fee there are (in simple terms about) continually 2 opportunities. in this party both opportunities are: -2x + 3 = x - 15 and -2x + 3 = - (x - 15) = -x + 15 fixing each and each and every , we get: x = 6 or x = -12

Answer 3

Again,

There are several dozen ways to interpret your problem, due to inconsistent spacing between monomials. Did you mean this?

( x^2 – 2x – 15 ) / ( x^2 + x – 6 ) / [ ( x^2 + x – 6 ) / ( x^2 – 2x – 35 ) ]

By the way, props for using the minus sign instead of the hyphen-minus.

Answer 4

I think, there are some errors. in the question
is the sign * or /

and there are two expressions as same


X^2 – 2x –15 x^2 + x –6
---------------- * ----------------
x^2 + x –6 x^2 –2x –35

=

(x+3)(x-5) (x-2)(x+3)
-----------------------------
(x-2)(x+3) (x+5)(x-7)

= (x+3)(x-5) / (x-7)

Answer 5

First, combine the equations

(X² - 2X-15) * (X²+x-6)
__________________
(X² + X - 6) * (X²-2x-35)

The (X²+x-6)'s cancel out so your new equation is

(X² - 2X-15)
_________
(X²-2x-35)


then factor those two equations

(X-5)(X+3)
_________
(X-7)(X+5)

this is the simplified answer

Hope I helped

Answer 6

after u simply .....u get (x + 3) (x - 5) divided by (x - 7) (x + 5)