## anonymous one year ago helpppppp

1. anonymous

@jjamz87

2. anonymous

ik its choice a and b but ineed the restriction?

3. DecentNabeel

$\frac{x^2+4x-45}{x^2+10x+9}=\frac{x-5}{x+1}$

4. mathstudent55

The first step is to factor the numerator and denominator.

5. anonymous

is the restriction -9

6. DecentNabeel

$\mathrm{Factor}\:x^2+4x-45:\quad \left(x+9\right)\left(x-5\right)$ $=\frac{\left(x+9\right)\left(x-5\right)}{x^2+10x+9}$ $\mathrm{Factor}\:x^2+10x+9:\quad \left(x+9\right)\left(x+1\right)$ $=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}$ $=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}$ $\mathrm{Cancel\:the\:common\:factors:}\:x+9$ $=\frac{x-5}{x+1}$ that is the answer

7. DecentNabeel

@jammy987 are you understand

8. triciaal

|dw:1439493910772:dw|

9. triciaal

you do not want factor to be 0 so restrictions are the values that will make the factor = 0

10. triciaal

restrictions x not = -1, not = 5, not = -9

11. triciaal

so option A is correct do not divide by 0

12. mathstudent55

You don't want the denominator to be zero. The numerator can be zero. These two factors cannot be zero: $$x + 9 \ne 0$$ $$x + 1 \ne 0$$ x - 5 .can be zero because a zero in the numerator is permitted.