Algebra II Hints

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.

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) = x2 -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)=2x2+4x-8, g(x)= 5x-3, h(x)=1/(x-2)
(f+g)(x) f(x)+g(x) = (2x2+4x-8) + (5x-3) = 2x2+4x-8+5x-3 = 2x2+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) = (2x2+4x-8) – (5x-3) = 2x2+4x-8-5x+3 = 2x2-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) = (2x2+4x-8).(5x-3) = 10x3+20x2-40x-6x2-12x+24
= 10x3+14x2-52x+24
(f.g)(-2) f(-2).g(-2) =
(f/g)(x) f(x)/g(x) = (2x2+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
(goh)(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)
(goh)(-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
(hog)(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 FOILFirst, Outer, Inner, and Last. For example:

(ax + b)(cx + d) = acx2+ adx + bcx + bd = acx2 + (ad + bc)x + bd

In most cases a and c are 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) = x2 + 5x+3x + 15 = x2 + (5+3)x + 15 = x2 + 8x + 15
(x + 3)(x - 5) = x2 - 5x+3x - 15 = x2 + (-5+3)x - 15 = x2 - 2x - 15
(x - 3)(x + 5) = x2 + 5x-3x - 15 = x2 + (5-3)x - 15 = x2 + 2x - 15
(x - 3)(x - 5) = x2 - 5x-3x + 15 = x2 - (5+3)x + 15 = x2 - 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         =        1    
 x2-9              (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/x) -1)(x + 2) = 24 miles
Multiplying both sides by x produces:
(24 - 1x)(x+2) =24x
Using FOIL:
24x + 48 - x2 - 2x = 24x
Combining:
-x2 + 22x + 48 = 24x
+x2 - 22x - 48 = +x2 - 22x - 48
0 = x2 +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