This section consists of Algebra II hints, based on common errors that I see on student tests. I also provide a helpful hints for Algebra I, which would also be helpful.

## Algebra I

These are the topics in the Algebra I page.

- Order of Operation
- Distribution
- Rates
- Percentage Problems
- Solutions to Solution Problems
- Trains Approaching
- Representation of Inequalities
- Set Builder Notation
- Interval Notation

## Algebra II

As common problems come up in Algebra II, I will add to this section as needed. The topics are as follows:

### Combining Function

Functions come in the general format of f(x) = function of x. I.E. f(x) = x^{2} -3x + 2. Functions can be combined or act on one another. There are specific rules for combining functions. Following are a few basic examples.

Function | How to Use |
---|---|

Functions | f(x)=2x^{2}+4x-8, g(x)= 5x-3, h(x)=1/(x-2) |

(f+g)(x) | f(x)+g(x) = (2x^{2}+4x-8) + (5x-3) = 2x^{2}+4x-8+5x-3 = 2x^{2}+9x-11 |

(f+g)(-2) | f(-2)+g(-2) = (2(-2)^{2}+4(-2)-8)+(5(-2)-3) = 2(-2)^{2}+4(-2)-8+5(-2)-3 = 2(4)+9(-2)-11=8-18-11=-21 |

(f-g)(x) | f(x)-g(x) = (2x^{2}+4x-8) – (5x-3) = 2x^{2}+4x-8-5x+3 = 2x^{2}-x-5 |

(f-g)(-2) | f(-2)-g(-2) = (2(-2)^{2}+4(-2)-8)-(5(-2)-3) = 2(-2)^{2}+4(-2)-8-5(-2)+3 = 2(4)-(-2)-5=8+2-5=5 |

(f^{.}g)(x) |
f(x)^{.}g(x) = (2x^{2}+4x-8)^{.}(5x-3) = 10x^{3}+20x^{2}-40x-6x^{2}-12x+24= 10x ^{3}+14x^{2}-52x+24 |

(f^{.}g)(-2) |
f(-2)^{.}g(-2) = |

(f/g)(x) | f(x)/g(x) = (2x^{2}+4x-8)/(5x-3) |

(f/g)(-2) | f(-2)/g(-2) = (2(-2)^{2}+4(-2)-8)/(5(-2)-3) = (8+4)/(-10-3) = 12/(-13) = -12/13 |

(g^{o}h)(x) |
(g(h(x)) = g(1/(x-2)) = 5(1/(x-2)) – 3 = 5/(x-2) – 3(x-2)/(x-2) = 5/(x – 2) – (3x – 6)/(x-2) = (5 – 3x + 6)/(x -2) = (-3x + 11)/(x-2) |

(g^{o}h)(-2) |
(g(h(x)) = g(1/(-2-2)) = 5(1/(-4)) – 3 = 5/(-4) – 3(-4)/(-4) = 5/(-4) – (-12)/(-4) = (5 + 12)/(-4) = (17)/(-4) = -4 ^{1}/_{4} |

(h^{o}g)(x) |
(h(g(x)) |

### FOIL

In multiplying expressions, every element of the first expression must be multiplied by every element of the second expression. Normally, this is a two by two multiplication and the way to remember how this is done is through **FOIL** – *First, Outer, Inner,* and *Last*. For example:

(ax + b)(cx + d) = acx^{2}+ adx + bcx + bd = acx^{2}+ (ad + bc)x + bd In most casesaandcare 1, so the middle expression becomes(b + d)x. In this case, the number in front of the x is the sum of the two number that when multiplied gives the final element of the resulting equation. For example:

(x + 3)(x + 5) = x^{2}+ 5x+3x + 15 = x^{2}+ (5+3)x + 15 = x^{2}+ 8x + 15 (x + 3)(x - 5) = x^{2}- 5x+3x - 15 = x^{2}+ (-5+3)x - 15 = x^{2}- 2x - 15 (x - 3)(x + 5) = x^{2}+ 5x-3x - 15 = x^{2}+ (5-3)x - 15 = x^{2}+ 2x - 15 (x - 3)(x - 5) = x^{2}- 5x-3x + 15 = x^{2}- (5+3)x + 15 = x^{2}- 8x + 15

### Factoring

Use FOIL in reverse to calculate the factors from a polynomial.

### Rational Expressions

A *Rational Expression* is a ratio, or fraction, that includes variables in the elements of the ratio. For example:

x - 3_{=}x - 3_{=}1x^{2}-9 (x-3)(x+3) (x+3)

#### Simplifying

Simplifying is finding common factors, like in the example above, and cancelling them out.

#### Restricted Values

Restricted Values are values not possible for the value of the variable. In the example above, you would set the denominator to **0** and then solve for** X.** The two possible solutions are given for before and after simplification:

Before: (x-3)(x+3)=0->(x-3)=0 or x=+3 and (x+3) = 0 or x=-3 After: (x+3) = 0 or x = -3

It is better to simplify first and then find the domain from -∞ through +∞, with the exception for the Restricted Values.

Before: (-∞, -3) ∪ (-3, 3) ∪ (3, ∞) After: (-∞, -3) ∪ (-3, ∞)

### Two Variable set of equations

As a rule of thumb, you need as many equations as you do variables in order to solve a problem. In the examples here, there will be two unknowns so we need to create two equations and solve for them.

#### Two person bike ride

If Josh and Janice go for a bike ride. They decide on a 24-mile route. Janice rides 2 miles/hour faster than Josh and finishes the course 1 hour sooner. What are their speeds and times. The variables are:

Time = Distance/Speed

Rider | Speed | Time | Time2 = Distance/Speed |
---|---|---|---|

Josh | x mph | y hours | 24 miles/x mph |

Janice | x+2 mph | y-1 hours | 24 miles/((x+2) mph) |

Distance = speed * time. Therefore, plugging in for **y** from the Josh line into the Janice equation, I have:

((24 hours)/(x mph) -1 hour)((x + 2) mph) = 24 miles Multiplying both sides by x produces: (24 - 1x)(x+2) =24x Using FOIL: 24x + 48 - x^{2}- 2x = 24x Combining: -x^{2}+ 22x + 48 = 24x +x^{2}- 22x - 48 = +x^{2}- 22x - 48 0 = x^{2}+2x -48 Factoring: (x + 8)(x - 6) = 0 Since speed cannot be a negative number, the only answer is: x = 6 mph y = 24 miles/6 mph = 4 hours

Then the final table is:

Rider | Speed | Time | Time2 = Distance/Speed |
---|---|---|---|

Josh | 6 mph | 4 hours | 24 miles/6 mph |

Janice | 8 mph | 3 hours | 24 miles/8 mph |