The document provides instructions for solving three math problems: 1) Finding the domain of a radical function by simplifying the numerator and denominator and determining values where the denominator is not equal to zero. 2) Completing the square in vertex form by adding and subtracting values from both sides of the equation before taking the square root. 3) Factoring a polynomial expression using the grouping method by separating terms in parentheses and factoring out common factors.

1. Dev Project

2. Problem 1: Domain of a
radical Function.
-First you focus on simplifying the
numerator to its simplest form.
- Look for the greatest common
factor in the equation. (An 8 can
be factored out of the numerator)
-Once the 8 is factored out of the
numerator, you can see that the
numerator can be further
simplified by finding the “p”
values.

3. Domain of a radical
Function… continued
-Now that the numerator and
the denominator are in it’s
simplest form you can look for
the domain.
The domain for the numerator alone would be, D: (
The denominators “p” value cannot be equal to pi, because
denominator can not be zero. Because of this you have to set the
equation as greater than zero and solve for “p” (shown below).
By adding “p” to both sides you
would get an equation that
looks like…

4. Domain of a radical
Function…Continued
So now you put it together to get the final domain for the equation as a whole.

5. Problem 2: Completing the
square in vertex form
Given the equation…
Start by adding
50 to both
sides…
Factor 14 out of the
equation…
Because 9/196 was added to the left side, you have to add the same value to the
left side. Which follows the rule that whatever you add to one side, must be
added to the other.
Divide both
sides by 14 to
get rid of the
coefficient.

6. Completing the square in
vertex form… continued
The equation now looks
like…
And can be further
simplified to…
You then take the
square root of both
sides to get rid of
the left sides
exponent.

7. Completing the square in
vertex form… continued
You are then left with…
To solve for “n” you
then add 13/14 to
each side.
The “n” values are
then equal to:

8. Problem 3: completing the
square
Given the problem…
Begin by subtracting 9
from both sides…
Factor out a 3 from
the equation…
Since you add 256 to one side you add the equivalent to the other side
(512) .

9. Completing the
square…continued
Divide both sides by three to
get rid of the coefficient on the
left side.
You are left with…
Which can be simplified to…
Then you take the square root of
both sides to get rid of the
coefficient on the left side.
It then becomes…

10. Completing the
square…continued
You then add 16 to both sides
to solve for the “b” values…

11. Factor by Grouping
Given the problem…
In order to factor
the equation you
section the equation
with parenthesis…
You can the factor a
“7n” out of the
entire equation.

12. Factor by
grouping…continued
Once “7n” is factored out
you are left with…
You can further simplify
the first set of
parenthesis…
Because the (10n-1) is in
both you write it as this…