I need to Divide this problem?

Question

x^2-x-6 / x+4 divided by x^2-9 / x^2 +3x-4

This is all one problem x^2-x-6 over x+4 divided by x^2-9 over x^2+3x-4

Answer 1

Remember that division is the same as multiplying by the reciprocal.

x^2-x-6 / x+4 divided by x^2-9 / x^2 +3x-4  ....Factor all the quantities, change division to multiplication and flip the second fraction over.

(x-3) (x+2) * (x+4) (x -1)

_______________________

(x+4) (x+3) (x - 3)

Then cancel the quantities that appear in both numerator and denominator.

solution :  [(x+2)(x - 1)] / (x + 3)

Good luck to you !

Answer 2

mm funn. this will take me a while:

so its: x^2 -x-6/x+4 over x^2-9/x^2+3x-4

well first you flip the second one to multiply

so it'd be: x^2-x-6/x+4 times x^2+3x-4/x^2-9

factor: x^2-x-6 = (x+2)(x-3) and that's over x+4

factor: x^2+3x-4 = (x-1)(x+4)

and that's over x^2-9 which = (x+3)(x-3)

sooo put that all together:

(x+2)(x-3)/x+4 times (x-1)(x+4)/ (x+3)(x-3)

cross cancel:

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

from there you multiply:

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

and i think you get:

x^2+x-2/x+3

Answer 3

= ([x² - x - 6]/[x + 4])/([x² - 9]/[x² + 3x - 4])

= ([{x + 2}{x - 3}]/[x + 4])/([{x + 3}{x - 3}]/[{x + 4}{x - 1}])

= ([{x + 2}{x - 3}]/[x + 4])([{x + 4}{x - 1}]/[{x + 3}{x - 3}])

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

Answer: ([x + 2][x - 1])/(x + 3)