Synthetic Division - March 5, 2014

Hello, Rhenz here.
Today we learned about Synthetic division, it's a faster and possibly easier way to divide polynomials , comparing to long division. Keep note, i'm bad at explaining, sorry in advance.

Key steps of Synthetic Division ( This is in your unit booklet,page 17 ) :

1. Arrange the coefficients of f(x) in order of descending powers of x 
   ( Write 0 as the coefficient for each missing power. )

2. After writing the divisor in the form x - a, use "a" to generate the second 
   and third rows of numbers as follows. Bring down  the first coefficient of the    dividend and multiply it by "a": then add the product to the second                coefficient of the dividend.

3. The last number in the third row of numbers is the remainder; the other           numbers in the third row are the coefficients of the quotient, which is of         degree 1 less than f(x).



Before I give out examples, let's quickly review which are which.

 

Ignore the numbers, just remember where divisor,quotient,dividend,and remainder are.

Ex1- Use synthetic division to divide 

         6x³ + x⁴ - 3x² -x + 8 by x - 1

Step 1 :  

     6x³ + x⁴ - 3x² - x + 8    =>      x⁴ + 6x³ - 3x² - x + 8

   Arrange the coefficients of f(x) in order of descending powers of x 

Step 2: 

x - a = x - 1 

Divisor is 1 ( if you plug in x-1 into x-a, you will get x+1, divisor is a )

Dividend is 1 6 -3 1 8 ( Coefficients and constant of x⁴ + 6x³ - 3x² - x + 8 )

 Bring down  the first coefficient of the dividend and multiply it by "a": then add the product to the second coefficient of the dividend.

1  1+6-3-1+8

+    -1-5 8-7
 ------------------
   1  5 -8 7 1

Step 3: 

 x⁴ + 6x³ - 3x² - x + 8  / x - 1        =>        1x³ + 5x² - 8x - 7 + (1/x-1)

Put what you get from step 2 to the coefficient of 
quotient( x⁴ + 6x³ - 3x² - x + 8), which is of degree 1 less than f(x).
        

Ex2 - Use Synthetic division to divide 

         -x⁴ - x³ + x² - x + 2  by x + 2

Step1:

Coefficient and Powers are already arrange in descending order on this one, so no need.

Step2:

x - a = x + 2 

Divisor is -2 ( if you plug in x+2 into x-a, you will get x-2, divisor is a )

Dividend is -1 -1 1 -1 2 ( Coefficients and constant of -x⁴ - x³ + x² - x + 2 )

 Bring down  the first coefficient of the dividend and multiply it by "a": then add the product to the second coefficient of the dividend.

-2  -1 -1 1 -1 2

  +      2-2 2-2
  -----------------
    -1  1 -1 1 0

Step 3: 

-x⁴ - x³ + x² - x + 2  / x + 2        =>          x³ + x² - x + 1 

Put what you get from step 2 to the coefficient of 
quotient( -x⁴ - x³ + x² - x + 2 ), which is of degree 1 less than f(x).
On this case, our remainder is 0, so we just leave it out.

Ex3 - Use Synthetic division to divide

         4x³ - 15x + 2 by x - 3

Step1: 

4x³ - 15x + 2 => 4x³ + 0x² - 15x + 2

Arrange the coefficients of f(x) in order of descending powers of x 

In this case, we have a missing power, write 0 as the coefficient for each missing power.

Step2:

x - a = x - 3 

Divisor is 3 ( if you plug in x+2 into x-a, you will get x+3, divisor is a )

Dividend is 4 0 -15 2 ( Coefficients and constant of 4x³ + 0x² - 15x + 2 )

 Bring down  the first coefficient of the dividend and multiply it by "a": then add the product to the second coefficient of the dividend.

3  4  0 -15  2

  +    12 36 63
  -----------------
    4  12 21 65



Step 3: 

4x³ + 0x² - 15x + 2  / x - 3        =>          4x² + 12x - 21 + (65  / x - 3)

Put what you get from step 2 to the coefficient of 
quotient( 4x³ + 0x² - 15x + 2  / x - 3 ), which is of degree 1 less than f(x).

Other things;

To find out if a quadratic quotient is factorable or not, use discriminant.

Discriminant = B² - 4AC

IF

Discriminant>0 : Two real solutions

Discriminant=0 : One real solution

Discriminant<0 : No real solution

Ex-  4x² + 12x - 21

a = 4

b = 12

c = 21

B² - 4AC   =>   12²  - 4(4)(-21) => 144 - 336 => -192

Discriminant is less than 0, so no real solution.

Assignment 3 is due on friday, so is our unit 2 test, good luck to all;