Operations with Rational Expressions

When simplifying rational expressions, what is a restriction?

• A value of a variable which makes the denominator equal to zero

• 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)
  

Tags

CCSS.HSA.APR.D.7

CCSS.HSA.CED.A.3

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

•  $\frac{30a^2bc}{60ac^3}$  

•  $\frac{a^2bc}{2ac^3}$  

•  $\frac{30a^2bc}{60ac^3}$  

•  $\frac{ab}{2c^2}$  

Tags

CCSS.HSA.APR.D.7

Write the following quotient in simplest form:

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

•  $8$  

•  $\frac{8p-1}{p+3}$  

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

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

Tags

CCSS.HSA.APR.D.6

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

Tags

CCSS.HSA.APR.D.7

Write this product in simplest form:

 $\frac{h^2+3h}{2}\cdot\frac{4}{h+3}$  

•  $\frac{4\left(h^2+3h\right)}{2\left(h+3\right)}$  

•  $\frac{2h^2+3h}{h+3}$  

•  $2h$  

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

Tags

CCSS.HSA.APR.D.7