Polynomial Graphs: End Behavior

Hello Lgene here , I am going to discuss the "end behavior" of polynomial graphs.

When you're graphing (or looking at a graph of) polynomials, it can help to already have an idea of what basic polynomial shapes look like. One of the aspects of this is "end behavior", and it's pretty easy.

Even Degree

As you can see, even-degree polynomials are either "up" on both ends (entering and then leaving the graphing "box" through the "top") or "down" on both ends (entering and then leaving through the "bottom"), depending on whether the polynomial has, respectively, a positive or negative leading coefficient. 

Odd Degree

Odd-degree polynomials have ends that head off in opposite directions. If they start "down" (entering the graphing "box" through the "bottom") and go "up" (leaving the graphing "box" through the "top"), they're positive polynomials; if they start "up" and go "down", they're negative polynomials.

*Things to remember*

            -Even degree Polynomial's ends, end in same direction.

            -Odd degree Polynomial's ends ,ends in opposite directions.

And...

Graphing High Degree Polynomials(i was supposed to post this last Thursday, but i didn't have a chance,  because Mr.P didn't accept me on the blog.. peace :P)

When factoring polynomial, one approach is to find one of the roots (which we will call r)

Of the polynomial, then factor out (x-r).The resulting polynomial has a smaller degree, and we can repeat the process again and again until we have the final result.

Example: X³-6x²-x+30=0

If r is the root, then r must be factor of 30. Therefore possible values for r are±1; ±2; ±3; ±5; ±6; ±10; ±15; ±30.

We can test whether a possible root is an actual root by plugging the value in and seeing if the expression becomes 0; 2 is not a root since 2³ – 6(2)²  - 2  +  30 = 8 -24 -2 + 30 = 12, but -2 is a root since (-2)³ – 6(-2)²  - (-2) +30 = - 8 - 24 +2 + 30 =0.

Note that if the root is -2, then the factor is (x+2); specially, pay attention to the change of the sign!

Then we calculate x³ – 6x² - x + 30 / x+ 2, and get x² - 8x + 15, with remainder of zero.(take a look on how to do SYNTETHIC DIVISION at Shubham's blog)

Now we consider the polynomial x² – 8x + 15; if r is a root, then r must be a factor of 15. That narrows it down to ±1; ±3; ±5; ±15.

As it turns out,  3 is a root since (3)² – 8(3) + 15 =9 - 24 +15 = 0.

Therefore we now calculate x² - 8x + 15 / x-3, and get x-5.

Since we divided first by (x - 2), then by (x – 3), and finally got (x-5), we conclude that x³ – 6x² –x + 30 = (x+2) (x-3) (x-5).

Lastly,

LONG DIVISION

SOLVING LONG DIVISIONS!!!

Hello everyone, my name is Shubham. The one who sits next to Vincent, the guy who picked me up today to scribe next. I told you to choose a number instead of choosing me, but you choose me. At that time i was going to kill you for choosing me, but never mind :)

 So today i am gonna discuss, what i have learned in Mr. P's class today. which is to solve Long Divisions. 

There are two methods of solving long divisions:-

1. The normal division method which is a bit time consuming.

2. Synthetic Division method which is quite easy and saves our time. 

The result of division of a polynomial in x, P(x), by a binomial of the form P(x) = Q(x)+ R 

                                                                                                                         x-a               x-a

                                                          

=> NORMAL DIVISION METHOD

Example: divide  8x²+9x+3x³ -2 by x+2

Steps:

  1. write the dividend and divisor polynomial in descending powers.

       3x³+8x²+3x

2. Divide the leading term of the dividend by first term.

       3x³  = 3x²

        x

3. Multiple the divisor by newly formed term of the quotient using the distributive law and subtract the result.

                                                           3x² + 2x -1                    

                                             x+2   ) 3x³+8x²+3x-2

                                                      -    3x³+6x²

                                                                   2x²+3x

                                                                   2x²+4x

                                                                          -x -2

                                                                          -x -2

                                                                                 0

  factors = (x+2)(3x-x)(x+1)

                             

=> SYNTHETIC DIVISION

    Example: Divide 3x⁴ - 5x² - 4x + 4  by x+2

     

   Solution: P(x)= 0  =>  x+3=0 =>  x= -2

                    x-a     

Now divide all the coefficients by 3. And if any coefficient of power of x is missing in the         followings the let that coefficient be ZERO

like in this example the coefficient of x³ is missing then let that coefficient of x³  be zero

                                                                             

                                      -2  ) 2  +0  -5  +4  +  4

                                           +      -4   8   -6    -4

                                               2  -4   3   -2     0

Factor => 2x³-4x² +3x -2

The key steps of synthetic division are as follows:-

1. Arrange the coefficient of f(x) in the order of descending power of x

2. After writing the divisor in the form x-a, use "a" to generate the 2nd and 3rd rows of number as follow

3. Bring down the 1st coefficient of the dividend and multiple by "a" 

4. Then add the product to the 2nd coefficient of dividend.Repeat the process until a product is added to the constant term of (x).

Well, those are the ways Mr. P taught us in class to solve these long divisions and I hope all the readers of this blog will like amd take help of it to solve the questions.:)

Fun with Factoring

Fun with Factoring!

Hello! My name`s Vincent. The one who sits next to Aron, the one who picked me to scribe today (This is your fault, you`re supposed to pick Shubham, you traitor. haha,just kidding!)

So in this blog, I`m going to discuss what I learned in Mr. P`s class, which is FACTORING.
There are a lot of ways to factor numbers, and one of them is finding the GCF(Greatest Common Factor)The GCF of two or more monomials is the greatest common factor that divide each of the monomials.
Example:
A. 7x² + 35x =                 7x(x + 5)

You can also FACTOR BY GROUPING, If a polynomial contains four or more items. It may be useful to group them together and factor out a common factor from each group.
Example:        
 B. 3x + 9y + 5xz + 15yz =      (3x + 9y) + (5xz + 15yz)
                              3(x + 3y) + 5z(x + 3y)
                            (3 + 5z) (x + 3y)
Another way of factoring is 
FACTORING THE DIFFERENCE OF TWO SQUARE
Sometimes an expression will be of the form (ax)² - b², which is similar to   a²b² - b². These expressions can be factored as follows: (ax)² - b² = (ax + b)(ax - b).  
Example:                
C. x² - 9 =                      (x)² - (3)²
                             (x + 3)(x - 3)

There`s also a way by FACTORING DIFFERENCE OF CUBES.Used for binomials that are a difference of two perfect cubes.
                        x³ - y ³ = (x - y)(x² + xy + y²)
Example:
D. x³ - 8 =                  (x)³  - ( 2 )³           
                       (x - 2)(x² + 2x + 4)       

You can also try FACTORING SUM OF CUBES.
Used for binomials that are a sum of two perfect cubes.
                         x ³ + y ³ = (x + y)(x² - xy + y²)
Example:
E. x³ + 27 =                  (x)³  + ( 3 )³           
                       (x + 3)(x² - 3x + 9) 

Another way of factoring is 
FACTORING PERFECT SQUARE TRINOMIALS.The trinomilas x² + 2xy + y² and   x² - 2xy + y² are called perfect square trinomials:
                    x² + 2xy + y² = (x + y)²
                    x² - 2xy + y² = (x - y)²
Example: 
F. 16x² + 24 + 9 =         (4x)² + 2 . 4x . 3 + (3)²                                              (4x + 3)²

Then there`s the FACTORING TRINOMIALS OF THE FORM x² + bx + c WITH LEADING COEFFICIENT 1
When given an expression of this form, we can break it into two binomials that can be multiplied together to get the given expression. 
x² + bx + c = (x + u)(x + v), where u and v are integers that satisfy the following:

1. uv = c
2. u + v = b

Here are some tips for this kind of factoring.
1. If c is (+) = u and v have the same sign.
    If b is (+) = u and v are (+)
    If b is (-) = u and v are (-)
2. If c is (-) = u and v have opposite signs.

Example: 
G. x²+ 7x + 12 =        a = 1, b = 7, c = 12
                            C > 0, B > 0
                              u.v = 12
                              u + v = 7 
                               3 , 4
                        (x + 4)(x + 3)




You can also use trial and error method which is a bit time consuming when your doing an exam.

Well, those are the factoring ways Mr. P taught us in class and I hope all the readers of this blog will like it.



Oh yeah since we haven't got donuts yet, here's something to motivate you :)