Can you simplify this rational expression?(: algebra1?

By factoring, we obtain:

(x - 7)(x - 5)/((x - 7)(x + 4))

Reduce the common factors of x - 7 to get:

(x - 5)/(x + 4)

I hope this helps!

for x^2-12x+35 find 2 numbers which the multiply is 35 and sum is -12 so we have -7 & -5

for x^2-3x-28 ind 2 numbers which the multply is -28 and sum is -3 so we have 4 & -7

(x-7)(x-5)

--------------

(x-7)(x+4)

so (x-7) will cancel the we will have

(x-5)

---------

(x-4)

first you factorize both the numerator and denominator. lets start with the numerator:

numerator: x^2 - 12x + 35

x^2 - 7x - 5x + 35

x(x - 7) -5 (x - 7)

(x - 5) (x - 7)

now lets factorize the denominator:

x^2 - 3x - 28

x^2 - 7x + 4x - 28

x(x - 7) + 4 (x - 7)

(x + 4) ( x - 7)

so we have:

(x - 5) (x - 7)/ (x + 4) (x - 7)

we cancell (x - 7) in the numerator and denominator and we get:

(x - 5)/ (x + 4)

x^2 - 12x + 35

--------------------

x^2 - 3x - 28

= (x -7)(x-5) / (x-7)(x+4)

= (x-5) / ( x+4)

(x^2 - 12x + 35) / (x^2 - 3x - 28)

=

(x-7)*(x-5)/[(x-7)(x+4)]

=

(x-5)/(x+4)

Ta da!

x^2 - 12x + 35

--------------------

x^2 - 3x - 28

(x-5)(x-7)

------------- =

(x+4)(x-7)

(x-5)

-------

(x+4)

x^2 - 12x + 35

---------------------

x^2 - 3x - 28

=

(x-7)(x-5)

------------

(x-7)(x+4)

=

(x-5)

------

(x+4)

(x²-12x+35)/(x²-12x-28)=[(x-5)(x-7)]/[(x-7)(x+4)]

=(x-5)/(x+4)

by factoring the top and bottom you get

(x - 7) (x - 5)

------------------ =

(x - 7) (x + 4)

you can then cancel the x-7 from top and bottom to get

x - 5

-------

x + 4

factorising top and bottom we get (x-7)(x-5)/(x-7)(x+4)

(x-7) cancel each other out giving (x-5)/(x+4)