# Polynomials

#### Cockchestner007

For the following rational expressions, perform the indicated operation and simplify the result:

a) y³-1 y-1
__________ ______
y² + y - 6 divided by 3+y

b) are there any restrictions on the variable y?

Hey guys i need help on how to do this, i have no idea where to start (Worried)

#### Cockchestner007

For the following rational expressions, perform the indicated operation and simplify the result:

a) y³-1 y-1
__________ ______
y² + y - 6 divided by 3+y

b) are there any restrictions on the variable y?

Hey guys i need help on how to do this, i have no idea where to start (Worried)
its y-1 / 3+y

#### masters

MHF Helper
For the following rational expressions, perform the indicated operation and simplify the result:

a) y³-1 y-1
__________ ______
y² + y - 6 divided by 3+y

b) are there any restrictions on the variable y?

Hey guys i need help on how to do this, i have no idea where to start (Worried)
Hi Cockchestner007,

I'm having a wee bit of trouble untangling what you have.
Is it this?

$$\displaystyle \frac{y^3-1}{y^2+y-6} \div \frac{y-1}{3+y}$$

If not, you might want to use grouping symbols to show exactly what you have.

#### Cockchestner007

yes!! thats exactly what i have

#### masters

MHF Helper
Hi Cockchestner007,

$$\displaystyle \frac{y^3-1}{y^2+y-6} \div \frac{y-1}{3+y}$$

Invert the divisor and multiply.

$$\displaystyle \frac{y^3-1}{y^2+y-6} \cdot \frac{y+3}{y-1}$$

$$\displaystyle \frac{(y-1)(y^2+y+1)}{(y+3)(y-2)} \cdot \frac{y+3}{y-1}$$

The "restrictions of the variable" means "what would y have to be to make the denominator of either fraction zero?"

Those restrictions would be: y cannot be -3, 2, or 1
Simplifying we arrive at:

$$\displaystyle \frac{y^2+y+1}{y-2}$$

The only restriction after simplification is y cannot be 2.
If we were talking about rational functions, this would be a vertical asymptote.

The other two restrictions would be points of discontinuity on the graph.

But, this may be too much information since these are only expressions.