Adding and Subtracting Rational Expressions

7.3 - Special Factoring

Difference of Squares:
x^2 - y^2 = (x + y) (x - y)
Perfect Square:
x^2 +2xy + y^2 = (x + y)^2
x^2 - 2xy + y^2 = (x - y)^2
Difference of Cubes:
x^3 - y^3 = (x + y)(x^2 + xy + y^2)
Sum of Cubes:
x^3 + y^3 = (x + y)(x^2 -xy +y^2)

7.1 , 7.2

7.1
Ex 1: Factor out greatest common factor.
1. 10x - 30
= 10(x+3)
Ex 2: Factor by grouping.
1. 2k + 2h + jk + jh
=2(k + h) j(k+h)
= (k + h)(2 + j)
7.2
EX 1: Factor the trinomial.
1. y^2 + 7y -30
= (y - 3) (y + 10)
EX 2:
1. 7p^2 +15pq +2q^2)
= (7p + ?) (p + ?)
= (7p + q) (p + 2q)

Remember Chapter 8!!!

So we also did an intro to functions in Chapter 8.

This includesRational Expressions and Functions: Multiplying and Dividing

And Adding and Subtracting Rational Expressions

For functions remember that f(x)=y and g(x)=y

How to pronounce: "f of x" of "g of x"

Final Review!

This is what we have learned this term in math!

Chapter 6: Exponents, Polynomials, and Polynomial Functions

6.1- Integer Exponents and Scientific Notation

6.2-Adding and Subtracting Polynomials

6.3-Polynomial Functions

6.4-Multiplying Polynomials

6.5 Dividing Polynomials

Chapter 7: Factoring

7.1-Greatest Common Factors; Factoring by Grouping

7.2-Factoring Trinomials

7.3-Special Factoring

7.4-Solving Equations by Factoring

Chapter 12: Nonlinear Functions, Conic Sections, and Nonlinear Systems

12.1-Additional Graphs and Logarithmic Functions

Chapter 11: Exponential and Logarithmic Functions

11.1-Inverse Functions

Chapter 10: Quadratic Functions, Inequalities, and Functions

10.1-The Square Root Property and Completing the Square

10.2-The Quadratic Formula

10.3-Equations in Quadratic Form

Square root Propery

If x and k are complex numbers and x^2 = k, then

x = SQRT k or x = -SQRT K

non quadratic function

A non quadratic function can be written like this: au^2 + bu + c = 0

Ways to solve quadratic equations

1. Factor--zero factor property
2. Square root property
3. Completing the square, difficult to factor trinomial

Remember

You cannot distribute an exponent over addition and subtraction.

Quadratic formula reminder

This is a great way to make sure that your know how to use the quadratic formula


http://www.youtube.com/watch?v=s80J2dAUUyI

Page 634

In class today, JoJo told us about page 634, a summary of all the importnat points gone over in 10.1-10.3. It's helpful because it gives you a quick rundown of all the important info that will be on our test tomorrow.
Good luck on your test!
Brigid

Important things to remember for the test

The Discriminant
If a, b and c are integers then the discriminant, b^2-4ac of ax^2+bx+c=0 determines the number and type of solutions as follows.
If the descriminate is posivtive the square of an integer then the number of solutions is two rational solutions.
If the discriminate is positive not the square of an integer then the numer of solutions is two irrational solutions
If the descriminate is zero then the number of solutions is one rational solution.
If the discriminate is negativwe then the number of solutions is two nonreal complex solutions.

10.3 Review

Here is a review problem from 10.3

x^4+x^2-12=0
(x^2)2 u=x^2
u^2+u-12
(u+4)(u-3)
u=-4 and u=3
x^2=-4 and x^2=3
x= SQRT-4 and x=SQRT3
x=2i and x=SQRT3

Test Review 10.1-10.3

Something that I found in the book that I think is helpful to go over for the test is on page 589 in our books. On the page there is a chart and it states all of the different methods for solving quadratic equations and each methods advantages and disadvantages. I think this is really helpful because it tells you when you should use which method and which method would be easiest to use when solving a certain equation! On the same page there are also some review problems that are helpful!

10.3 Equations in Quadratic Form Examples

Ex1:
X^4-13x^2+36=0
(x^2)^2-=363x^2+36=0 => u=x^2
u^2-13u+36=0
(u-9)(u-4)
UX^2=9 or x^2=4
x=+/-3 or x=+/-2
x= 3,-3,2,-2
Ex2:
2(4m-3)^2+7(4m-3)+5=0
4m-3=p=>2p^2+7p+5=0
(2p+5)(p+1)=0
p=-5/2 or p=-1
4m-3=5/2 or 4m-3=-1
4m= -5/2+3
4m=1/2
m=1/8
m=1/2
m= 1/8,1/2
m=1/8

when solving equations that begin in fraction form and then are changed through multiplying by a LCD into quadratic form you must always state your domain restrictions to ensure that you aren't dividing by zero.

10.3 solving equations in quadratic form

2(4m-3)^2+7(4m-3)+5=0
2u^2+7u+5=0 let 4m-3=u
(2u+5)(u+1)=0 factor zero factor prop.
2u+5=0 or u+1=0


u=-5/2 or u=-1

4m-3=-5/2 or 4m-3=-1 substitute 4m-3 for u

4m=1/2 or 4m=2 solve for m

m=1/8 or m=1/2

More on 10.3!

k= SQRT of 6k-8

-take the square of both sides.

k^2=6k-8

k^2-6k+8=0

(k-4)(k-2)=0

k=4 or k=2

-to check, plug into ORIGINAL equation.

x+ SQRT of x=6

-isolate the square root term.

-"work our magic" with squaring terms.

SQRT of x=6-x

x=(6-x)^2

x=(6-x)(6-x)

x=36-12x+x^2

-x -x

x^2-13x+36=0

(x-9)(x-4)=0

x=9 or x=4

when checking, 9 doesn't work so 4 is the only solution.

10.3

1/x+1/x-1=7/12

1. Clear fractions using LCD.

LCD=12(x)(x-1)

12(x)(x-1) over x + 12(x)(x-1) = 7(12(x)(x-1)

12(x-1)+12x=7x(x-1)

12x-12+12x=7x^2-7x

24x-12=7x^2-7x

2. Put into quadratic form: ax^2+bx+c=0

7x^2-31x+12=0

(7x-3)(x-4)=0

7x-3=0 or x-4=0

x=3/7 or x=4

3. You could...

factor

complete the square

quadratic form

square root property

*you must show the domain restrictions!

x cannot equal o or 1

Ok so on problem 27 in the hw for tonight the answer is -2+_ sr i2/3
when I solved the problem I got -2+_ sr i19 / 3
as you can see the only thing I had simplified differently was the 38, can some on please help me

The Discriminant

"b^2-4ac" > 0 2 distinct real solutions
"b^2-4ac" = 0 1 real solution (double root)
"b^2-4ac" < 0  0 real solutions

Imaginary Numbers!

-The square root of a negative number is imaginary, (i)

-The square root of -1 is i.

How you write i as an answer, Ex. 12i, not i12.

Ex. 9q^2-6q+5=0

has no real solutions but has two imaginary solutions.

10.3 Equations Quadratic in form

7x^2-31x+12 Standard form

(7x-3)(x-4)=0 Factor

7x-3=0 or x-4=0 zero-factor property

x=3/7 or x=4 solve each equation


Check by substituting in original equation

Ike's Panera Bread Fun!

"In May 1999, all of Au Bon Pain Co., Inc.'s business units were sold, with the exception of Panera Bread, and the company was renamed Panera Bread. Since those transactions were completed, the company's stock has grown thirteen-fold and over $1 billion in shareholder value has been created. Panera Bread has been recognized as one of Business Week's "100 Hot Growth Companies." As reported by The Wall St. Journal's Shareholder Scorecard in 2006, Panera Bread was recognized as the top performer in the restaurant category for one-, five- and ten-year returns to shareholders."

HAYO!

http://www.panerabread.com/about/company/history.php

Discriminant

(b^2-4ac) tells the number of REAL and/or UNIQUE solutions.

b^2-4ac > 0, we have 2 distinct, real solutions

b^2-4ac = 0, there is 1 real solution (double root)

b^2-4ac < 0, there are 0 real solutions

Ex. Find the discriminant of each equation, then, determine the number of real solutions.

a. 2x^2+4x+1=0

ax^2+bx+c

a=2,b=4,c=1

(4)^2-4(2)(1)

16-8=8

2 real solutions.

b. 2x^2+4x+2=0

ax^2+bx+c

a=2,b=4,c=2

(4)^2-4(2)(2)

16-16=0

There is 1 real solution

c. 2x^2+4x+3=0

ax^2+bx=c

a=2,b=4,c=3

(4)^2-4(2)(3)

16-24=-8

There are 0 real solutions

Flickr Diary -Need Points?

Flickr Diary


This is your chance to tell a algebraic graphic story.  You should use pictures that are bold and speak to the viewer.  Add a caption below your picture. More instructions below!


Tag all photos a2a11. Additional tags are listed in parenthesis. If no, parenthesis then tag using word next to the number.  Also tag with your name.

Also, note how each picture visual associates concepts learned for you personally.

Due by the final exam period.

1. Rectangular Coordinate Grid (PiGrid)
2. A Parabola (PiParab)
3. Polynomial (PiPoly)
4. Exponents (PiExp)
5. Real life representation -- graph of Function (PiFunction)
6. Real life symmetry (PiSymmetry) -spell correctly!
7. PiReflect
8. PiMonomial
9. Common Factor (PiCommon)
10. piquadratic
11. PiRatio
12. PiProportion
13. PiFunction
14. PiFunctionMachine
15. PiLine
16. PiAllometric
17. PiRadicalFunction
18. PiAbsvalue
19. PiCubicfuncgraph
20. Functions in Athletics (PiSport)
21. Piinequality
22. Piprojectile

Quadratic Formula

-b+/- SQRTb^2-4ac

2a

-There are two solutions because it can either be negative or positive

-Solutions same as roots

-zeros and or x-intercepts

Ex. x^2+5x-14=0

a=1, b=5, c=-14

x=-b-5+/-SQRT5^2-4(1)(-14)

2a

-5 +/- SQRT 25-(-56)

2

= -5 +/- SQRT 81

2

=5+/- 9

2

x=2 and -7 = {2,-7}

Rationalize the Denominator

-a solution is not simplified if there is an irrational number in the denominator of a fraction.

Ex. The square root of 121/7, the same thing is the square root of 121 over the square root of 7.

=11/the square root of 7

Next, multiply this fraction by the square root of 7 over the square root of 7, this equals 1.

=11 SQRT 7/7

Now, the denominator has a rational number, 7.

10.1 example equations

X^2=-17
√17, √-17

(k+5)^2=-100
√-100
-5+√10
-5,-10

2r^2-4r+1=0
2r^2+4r=-1x4
-2x2=4

3z^2-6z-2-0
3z^2=2+9
z=3+√11/3

HW problem #35

Some people had trouble with this problem.
Here it is step by step
#35

3w^2-w-24=0
3w^2-w=24
3(w^2-1/3w+(-1/6)^2)=24 +3(-1/6)^2
3(w^2-1/3w+1/36)=24+3/36
3(w-1/6)(w-1/6)=24+3/36
-1/6+-1/6=1/36
-1/6+-1/6=-2/6=-1/3
root(w-1/6)^2= root 289/36
w-1/6= = +- root 289/ root 36
w-1/6= +- 17/6
w=1/6 + 17/6 =18/6=3 OR -16/6= -8/3
{3, -8/3}

Completing the Square

Completing the Square

-forces any quadratic expression to factor

-use “completing the square” to solve quadratic equations

To Complete the Square:

Ex. x^2+6x+16=0

Note: a=1, b=6, c=-16

Step 1. Move the constant term ( c ) to one side and every other term to the other side.

x^2+6x-16=0 → x^2+6x=16

Step 2: Factor out the leading coefficient (a)

*will be more relevant when a does not equal 1.

1(x^2+6x)=16

Step 3: Add (1/2 x b)^2 to side with (a) and (b) term. Then add a(1/2 x b)^2 to the side with constant term.

Simplify:

1(x^2+6x+(6/2)^2)= 16+1(6/2)^2

x^2+6x+9=16+9

x^2+6x+9=25

Step 4: Factor side with (a) and (b) terms

(x+3)(x+3)=25 → (x+3)^2=25

*At this point you can either solve for x

Something to watch out for

If the solution is less than or equal to zero then using the square root property always produces two square roots one positive and one negative.

10,1 Completing the Square

Simple steps to Completing the Square
To solve ax^2+bx+c=0 (a greater than or equal to zero)
Steps
Step 1: Be sure the squared term has a coefficient 1. If the coefficient of the squared term is some other nonzero number a, divide each side of the equation by a.
Step 2: Write the equation in the correct form so that the terms with variables are on one side of the equals sigh and the constant is on the other side.
Step 3: Square half the coefficient of the first-degree term.
Step 4: Add the square to each side.
Step 5: Factor the perfect square trinomial. One side should now be a perfect square trinomial. Factor it as the square of a binomial. Simplify the other side.
Step 6: Solve the equation. Apply the square root property to complete the solution.

10.1!

10.1 The Square Root Property and Completing the Square

-the square root property allows you to eliminate the squared property of a number.

a means, “what two of the exact same number can be multiplied to get a?”

Ex. x^2=36

x^2= 36

x= 6 and x=-6

Because 6x6=36 and -6x-6=36

Standard Form of Quadratic Equation

ax^2+bx+c=0

A quadratic equation is a second degree equation because the term with the largest sum of exponents is ax^2 and its exponent is 2.

Ex. 4(x^2)+(-48x)+8=0

2(x62)+36x+6=0

Zero Factor Property

That is, if ab=0, then a=0 or b=0.

(3x+7)(x-4)=0

(a) x (b)=0

Either quantity a has to equal zero or quantity b has to equal zero in order for this statement to be true.



Properties of Square Roots, only used if a is greater than or equal to zero or b is greater than or equal to zero.


Ex. Simplify the following radical.
20 *Find the largest perfect square that goes into 20 evenly.

20=
4x5= 4x5= 25

Solve a quadratic equation in the form (ax^2+b)^2=c

Ex. (x-5)^2=36
(x-5)2= 36
((x-5)2)= 36
(x-5) =6 or (x-5)=-6
x=11 or x=-1
0 the square root could be positive or negative.

Verticle and Horizontal line test

Verticle line test: -Functions are only functions if they pass the verticle line test. The verticle line test is when you run a verticle line over a graph and see if in any places on the graph, the line hits more than one point. Horizontal Line test: You use the horizontal line test to see if a function is one-to-one. Similar to the vertical line test, you run a horizontal line across the graph and if the line gets to a point where it hits two points, then the funciton isn't one-to-one.

Transformations of graphs

Test Review Transformations of graphs: f(x) = - a (x +/- h) +/- k - The negative sign is the reflection a - stretches or flattens +/- h - Shifts left or right (opposite of sign) +/- k - Shifts up or down (same as sign)

Inverse Equation

I was confused in class today about the whole switching of variables when you use the inverse equation. If you think about it like the original variables are in the top and bottom equations and the switched are in the two middle equations it gets much easier.

Extra Help

Remember when a number is on the inside of parentheses it is moved to the same number of units to the opposite side.  If it's a positive number it shifts to the left and if it's a negative number it shifts to the right.  

If  the number is on the outside of parentheses, you move it up if it's a positive number and down if it's a negative number.

easy way to remember composition of functions

when you think about f(g(x)) it can sometimes be confusing which part to work on first, especially if the variable is left as x, no number substituted. When I looked back over all my notes I saw the f of g of x and it made so much more sense to think about it that way, like working from right to left

Remember

The domain of the original function is the range of its inverse function.  The range of the original function is the domain of its inverse function.

vertical and horizontal line tests

The vertical line test proves if the graph is a function.  The horizontal line test proves whether the function is one to one.

Functions

Here is a link that should help with the test if you are having a little trouble

http://www.youtube.com/user/khanacademy#p/c/7AF1C14AF1B05894/39/VhokQhjl5t0

composition functions

here is an explanation of composition functions

http://www.youtube.com/watch?v=S4AEZElTPDo&feature=related

1 to 1

If the graph doesnt pass the horizontal line test then it is not 1 to 1

Inverse Functions

f(x)= Ix-4I-3

x f(x)
-2 3
-1 2
0 1
1 0
2 1

Inverse: [(3,-2),(2,-1),(1,0),(0,1),(1,2)]
the domain comes the range and the range becomes the domain in inverse functions.

f(x)=Ix-4I-3 GRAPH

-the inverse of a functions graph reflects upon the y=x

g(x)=2x+5
1. Replace function name with y, y=2x+5
2. Switch x and y x=2y+5
3.Solve for y=
y=x-5/2
4. Rename ywith inverse function name g(x)=x-5/2=1/2x-5/2

-If the original function is a function and the inverse of that is a function it is called an original function one to one.

Inverse Functions

If f is a function from a set A to a set B, an inverse function is from B to A. If input x produces an output y then an inverse function would be input y and output x
Domain-all possible set of input usually (x) values
Range-all possible set of outpt usually y values
Purplemath discusses what weere goin over in class today which is how to find a function's domain and range
http://www.purplemath.com/modules/fcns2.htm

Vertical line test

You also know if a set of coordinates is a function if it passes the vertical line test.  Draw a vertical (straight up and down) line through a line and it only intersects at one point, then it's a function.

On the daily quiz to day many of us forgot how to factor out problems like...
(x+3)^3. All you do to solve these problems is set up the problem so it looks like (x+3) (x+3) (x+3), next you factor the first (x+3) in to the second (x+3) giving you (x+3) (x^2+6x+9). Than distribute the last (x+3) throughout.
x^3+6x^2+9x+3x^2+18x+27
combine like terms and your answer is...
x^3+9x^2+27x+27

More on Composition Functions

Composition Functions

-when you apply a function rule to the result of another function rule you compose a function.

-you MUST know the notation to compose.

Notation: f(g)(x)= f(g(x))
pronounce- “f of g of x”
f composed of g composed of x.

-all functions are composed of x, x is the input.

f(g(x)) is the same as (fog)(x)

f(x)= x^2+4 g(x)= 2x+3

now instead of putting x as the input, put g(x).
f(g(x))=(2x+3) then apply all of the x values from the x problem
f(g(x))= (2x+3)^2+4
simplify
=(2x+3)(2x+3)+4
=4x^2+6x+9+4
f(g(x))= 4x^2+12x+13

Ex1: Find (gof)(2)
=g(f(2))
1. Always work inside out by plugging the input values in.
=g(f(2))= g(8)

Test Review!

Here is some review for the test on friday! So far we have done:

-Re-Introduction to Functions 

If you want more practice on this, you should go to the websites that we had some homework problems on that is posted on the blog and look at the slideshow again. The homework problems are really helpful because they also show the answers!

-12.1 Graphs of Functions and Composition Functions

To review this section it would be helpful to study all of the graph parent functions and review what other people have posted on the blog to help remember all of the different graphs. To review composition functions it would be helpful to do problems 7-10 in the chapter review (p.775). 

-11.1 Inverse Functions

We will probably go over this today in class, but it would be helpful to look at the notes we took on this section for homework last night. 

Parabola

y=ax^2+bx+c

y=a(x-h)^2+k

It opens up if a >0
down if a <0
The vertex is (h,k)

It has a x^2 term not y squared

functions

A function is one to one when each x-value corresponds to only one y-value and each y-value corresponds to just one x-value.

One-to-One Functions

One-to-one functions come up in section 11.1. A one-to-one function is a function where each x-value corresponds to only one y-value, and each y-value corresponds to only one x-value.

An example of this is:

a. (1,2), (3,8), (9,0)

This IS a one-to-one function because each x and y-value only appears once throughout the ordered pairs.

b. (4,7), (8,6), (4,3)

This IS NOT a one-to-one function because the 4 shows up throughout the ordered pairs twice as the x-value.

Helpful website for conic sections

http://www.intmath.com/plane-analytic-geometry/conic-sections-summary.php
General Form of the Circle

An equation which can be written in the following form (with constants D, E, F) represents a circle:

x2 + y2 + Dx + Ey + F = 0
Formal Definition A circle is the locus of points that are equidistant from a fixed point (the center).
Conic Section



If we slice one of the cones with a plane at right angles to the axis of the cone, the shape formed is a circle.

While doing the HW problems today I forgot that terms like (x+5)^2 are written as (x+5)(x+5) - I was writing the problem as x^2+25. This is a small but critical error that could make all the difference on the up coming test.
That's all...Weeeeerd Up!

Composition Functions

When you apply a function rule to the result of another function rule you compose a function.

-you MUST know the notation to compose.

Notation: f(g)(x)= f(g(x))
pronounce- “f of g of x”
f composed of g composed of x.

-all functions are composed of x, x is the input.

f(g(x)) is the same as (fog)(x)

f(x)= x^2+4 g(x)= 2x+3

now instead of putting x as the input, put g(x).
f(g(x))=(2x+3) then apply all of the x values from the x problem
f(g(x))= (2x+3)^2+4
simplify
=(2x+3)(2x+3)+4
=4x^2+6x+9+4
f(g(x))= 4x^2+12x+13

Ex1: Find (gof)(2)
=g(f(2))
1. Always work inside out by plugging the input values in.
=g(f(2))= g(8)

12.1 functions

f(x)=x^2- the squaring function

f(x)=|x|- absolute value function

f(x)=1/x the reciprocal function

f(x)=√ x square root function

12.1 exercises HW

for problems 5-18 you can put the function in your caculator and see what the graph is supposed to look like. It really helps

Domain and Range

The domain of an absolute value function is (-inf.,inf) and its range is [0,inf.)

The reciprocal function the domain and range are both (-inf.,0) or (0,inf.)

The square root function the domain is [0,inf.) and the range is [0,inf.)

Every Transformation

F(X)= -afunc(x+/- h)+/-K

(reflection (shift left (Shift up or

over x-axis) or right down same as

opp. sign) sign)

Every Transformation

Every Transformation

-the standard form for any function is:
f(x)=-a(x-h)+k

f(x)= - a func (x +/- h) +/- k

- =reflection

a= stretches/flattens

+/- h= shifts left or right

+/- k= shifts up or down.

f(x)=2x^2+3, parabola y=x^2

q(x)= 4x^3, cubic, y=x^3

g(x)=1/(x-3), allometric, 1/x=y

w(x)=x+27 -6, square root, yx

h(x)=-(x-3), absolute value, y=x

Types of Functions and what they're called.

f(x)=2x^2 + 3, parabola ,y=x^2
g(x)= 4x^3,cubic ,y=x^3
g(x)=1(x-3)=allometric, 1x=y
w(x)= (x+27)-6, square root ,y=x
h(x)= -x-3 ,abs. value, y=x

Wednesday, March 30, 2011

Hey, I found a website with images of graphed functions with all different degrees.
Also, remember the graph names for functions with x^2, Parabolic
And the function with 1 over X is Allometric
You guys should take a look, just click it, they're only graphs, hehe, no reading.[kinda]
http://library.thinkquest.org/2647/algebra/functype.htm

New Vocab

Abscissa = The X coordinate. The distance from a point to the vertical or y axis measured parallel to the horizontal or X axis.
Ordinate = nThe perpindicular distance of p from the x axis. The values of x and y together written x, y.

Vertical Line Test

Vertical line test if a vertical line passes through more than one point of the graph. then the relation is not a function. A parabola is an example of a function because if you draw a line through the graph it doesnt go through more than one point.

Re-Introduction to Functions Presentation

Re-Introduction to Functions

 


Assignment 1 Homework Links


1.  http://www.intmath.com/functions-and-graphs/1-introduction-to-functions.php


2.  http://www.intmath.com/functions-and-graphs/2b-functions-from-verbal-statements.php


Assignment 2 Homework Links


1.  http://www.intmath.com/functions-and-graphs/3-rectangular-coordinates.php

example

f(x) = 4x + 7x, g(x) = 7y + 2x

f(10) = 4(10) + 7(10), f(10) = 40 + 70, f(10) = 110

g(7) = 7(7) + 2(10) = 49 + 20 = 69

HW

what were the links for the homework?

Today we talked about real life examples of functions, so here's a few to help you guys grasp the concept.

The amount of air pollution is dependent on the number of cars on the road.

The distance a baseball is hit depends on the force it is struck with.

A person's hunger is dependent on the the amount of food they have recently consumed.

( time is always the independent variable because nothing can control time.)

Function notes

A function is a rule that relates how one quantity depends on other quantitites

V=IR where

V=Voltage (V)

I=current (A)

R= resistance

s=speed (m/s)

t=time take (s)

If d increases the speed goes up


S=1mi/t t is independent

y=x^2 + 3 time is on the x- axis and independent variable

independent varible represents all possible values of domain

function f of x...f(x)

Monday, March 28, 2011

Remember to NOT MULTIPLY!
These mathematical statements all mean the same!
Linear Equation:
y = 2x +3

Linear Function:
f(x) = 2x + 3

g(x) = 2x + 3

h(a) = 2a + 3

Chap 8. Rational Expressions

A rational number is the quotient of two integers, with the denominator not 0. In algebra a rational expression is the quotient of two polynomials with the denominator 0

x/y -a/y m+4/m-2 8x^2-2x+5/4x^2+5x

Just vocab...Expanding is the same thing as Foiling.

7.3 Special Factoring

Difference of Squares

x^2-y^2=(x+y)(x-y)


Perfect Square Trinomials
x^2+2xy+y^2=(x+y)^2
x^2-2xy+y^2=(x-y)^2

Difference of Cubes
x^3-y^3=(x-y)(x^2+xy+y^2)

7.2 Factoring Trinomials

Choose factors of the first term and factors of the last term. The place them in a pair of parenthesis

( )( )

Use different combinations of the factors until the correct middle term is found

Factoring out a Binomial factor

(x-5)(x+6)+(x-5)(2x+5)
the greatest common factor is (x-5)
(x-5)[(x+6)+(2x+5)]
commutative prop. of +
(x-5)(x+6+2x+5)
combine like terms
(x-5)(3x+11)

and that's how you factor out a binomial factor

More on X Method

When the polynomials first term is greater than one this is how you factor it:
You do all of the same steps as the first example.
1. Then when you rewrite the answer, you put the first term in the original problem, and divide it by the second term after you factored it.
2. Then you simplify the problem.
3. Next, if you have a fraction, then you take the denominator and multiply it by the first term(in the factored out problem)
4. That is your answer.

Factoring X Method

For factoring when the first term is 1:
You put the first and last term in the bottom of X and multiply them. You put the middle term in the top of the x. You find two numbers that add to get the middle term, and multiply to get the first and last term multiplied together. You put those numbers in the two sides of the x. Then you rewrite the new problem and that is the answer.

Factoring a Polynomial

These steps are listed int he book on page 393 and they help so remember to use these on the test tomorrow

Step1 Factor out any common factor
Step 2 If the polynomial is a binomial, check to see if it is the difference of sqaures
Step 3 If the polynomial is a trinomial , check to see if it is a perfect sqaure trinomial
Step 4 If the polynomial has more than 3 terms, try to factor the grouping
Final Step- Check the factored form by multiplying

Don't forget...

Don't forget to look over some word problems, 2nd period went over one in class and it seems like they could get a little confusing. some practice problems are on 403. remember to keep track of all of your variables and details!
good luck tomorrow!

Extra help

hey!so as I am studying for our test tomorrow I found that it is really helpful to do the chapter 7 test, it has a few different variations of problems that we didn't see on the review.

It also has a lot of factoring by grouping which we worked a lot in class.

Reaaaaally Easy Way to Factor

While I was studying, I was a little bit confused on how to use the AC Method. I found this amazing site that does the method a little differently but makes it so much easier!

http://people.richland.edu/james/misc/acmeth.html

chap 7 review nots

AC METHOD

6X^2-14X+9X-21 6x^2-5x-21 expanding (3x-7)(2x+3) factoring
2X(3X-7)+3(X-7)
(3x-7)(2x+3)


-126
/ \
-14 9

Factoring a Polynomial

Step 1: Factor out any common factors.

Step 2: If the polynomial is a binomial, check to see if there is a difference of squares.
If the polynomial is a trinomial, check to see if its a perfect square trinomial. If its not factor it using the methods we have learned.
If the polynomial has more than three terms, try to factor by grouping.

Step 3: Check your answer

Types of Factoring

Types of factors you can use on the test:
Factoring by Grouping
Guess and Check
Ac Method
X-Method
Box Method

Factoring a polynomial

Steps for factoring:

1) Factor out GCF

2) Check to see if the polynomial is a difference of squares

3) If the polynomial is a trinomial check to see if it's a perfect square trinomial.

4) If the polynomial has more than 3 terms try to factor by grouping

5) Check the factored form by multiplying

Solving Quadratic Equations by Factoring

Here are the steps to solving quadratic equations
 #1 Write in standard form.
Standard form= ax^2+bx+c=0
a is not equal to 0
#2  Factor the polynomial
#3 Use the zero-factor property. Set each variable factor eqaul to 0
#4 Find the soultion or solutions- Solve each equation
#5 Check each solution in the original equation

purple math is always god site to go to if you still have trouble
http://www.purplemath.com/modules/solvquad.htm

Special Types of Factoring

You need to know these special types of factoring for the test on Tuesday

Difference of Squares- x^2-y^2=(x+y)(x-y)
Perfect Sqaures Trinomial- x^2+2xy+y^2=(x+y)^2  or  x^2-2xy+y^2=(x-y)^2
Difference of Cubes- x^3-y^3=(x-y)(x^2+xy+y^2)
Sum of Cubes- x^3+y^3=(x+y)(x^2-xy+y^2)