When simplifying rational expressions, what is a restriction?

• A value of a variable that makes the expression mathematically illegal.

• A vertical asymptote

• A hole in the graph

• The negative or opposite of the denominator

• The negative or opposite of the numerator

How do we find the restrictions of a rational expression?

• We set the numerator equal to zero and solve for the variable

• We set the denominator equal to zero and solve for the variable

• We plug zero into the numerator and simplify

• We plug zero into the denominator and simplify

What are the restrictions of:    $\frac{c-5}{c+5}$  

•  $c\ne5$  

•  $c\ne-5$  

•  $c\ne5$  and  $c\ne-5$  

• (-5, 0)
  

Write this product in simplest form:
  $\frac{6a^2b}{15c^3}\cdot\frac{5c}{4a}$  

•  $\frac{30a^2bc}{60ac^3}$  

•  $\frac{a^2bc}{2ac^3}$  

•  $\frac{30a^2bc}{60ac^3}$  

•  $\frac{ab}{2c^2}$  

What are the restrictions of:    $\frac{8x}{\left(x+4\right)\left(x-5\right)}$  

(Pick all that apply!)

•  $x\ne5$  

•  $x\ne-5$  

•  $x\ne0$  

•  $x\ne4$  

•  $x\ne-4$  

Write the following sum in simplest form:

 $\frac{9x}{21}+\frac{x+7}{7}$  

•  $12x$  

•  $\frac{12x+7}{21}$  

•  $\frac{12x+21}{21}$  

•  $\frac{84x+147}{147}$  

Write the following quotient in simplest form:

 $\frac{8\left(p-1\right)}{p+3}\div\frac{p-1}{p+3}$  

•  $8$  

•  $\frac{8p-1}{p+3}$  

•  $\frac{8\left(p-1\right)\left(p+3\right)}{\left(p+3\right)\left(p-1\right)}$  

•  $\frac{8\left(p-1\right)\left(p+3\right)\ +\ \left(p-1\right)\left(p+3\right)}{\left(p+3\right)\left(p-1\right)}$  

What are the restrictions of:   $\frac{19}{8u\left(u+2\right)}$  

(Pick all that apply!)

•  $u\ \ne0$  

•  $u\ \ne2$  

•  $u\ \ne-2$  

•  $u\ne8$  

•  $u\ne-8$  

Write the following difference in simplest form:

 $\frac{a}{a+3}-\frac{2a}{3}$  

•  $\frac{-2a^2+9a}{3\left(a+3\right)}$  

•  $\frac{2a^2+9a}{3\left(a+3\right)}$  

•  $\frac{-2a^2-3a}{3\left(a+3\right)}$  

•  $\frac{\left(a-a^2\right)}{3\left(a+3\right)}$  

Student 1 says that the restrictions of $\frac{9\left(d-4\right)}{\left(d-7\right)\left(d+8\right)}$ are $d\ne4$ , $d\ne7$ , and $d\ne-8$ . Is Student 1 correct? Why?

• Yes because d = 4, d = 7, and d = -8 all make the denominator equal to zero

• Yes because d = 4, d = 7, and d = -8 all make the expression evaluate to zero

• No because d = 4 does not make the denominator equal to zero

• No because the real restrictions are $d\ne-4$ , $d\ne-7$ , and $d\ne8$

• No because d = 0 is also a restriction