+0  
 
+1
626
2
avatar+693 

polynomials 

 

http://prntscr.com/kz2fnp

 

http://prntscr.com/kz2fxb

 Sep 26, 2018
 #1
avatar+128407 
+2

First one

 

1x^4  - 4x^3 -22x^2 + 4x + 21

 

First of all...if we can add the coefficients and the constant at the end and get 0, then  1 is a root ...and this is true

 

So...we can use some sythetic division to find the other  roots

 

 

1 [  1     - 4      - 22       4       21]

               1        -3       -25    -21

     _______________________

      1      -3       - 25     -21     0

 

The remaining polynomial is   x^3  - 3x^2  - 25x  - 21

 

By the Factor Theorem   the remaining  possible zeroes  are ±1, ±3, ±7  or ±21

 

We can test these values in the polynomial....those that result in  "0"  means that that integer is a root

-1 is a root

-3 is a root 

7 is a root

 

We can stop here....a 4th power polynomial can't have more than  4 roots

 

So...the roots are   x = { -3, -1, 1 , 7 }

 

 

cool cool cool

 Sep 26, 2018
 #2
avatar+128407 
+2

Second one

 

x^4 - x^3 -7x^2 + x + 6

 

Since 3 is a zero, ( x - 3)  is one linear factor...we can use synthetic division to find the remaining polynomial

 

 

3 [  1     - 1       - 7      1        6  ]

               3        6      -3      - 6

     _____________________

     1       2        -1     -2         0

 

The remaining polynomial is

 

x^3  + 2x^2  - x  - 2       we can factor this as

 

x^2 (x + 2)  - 1 (x + 2)   =

 

(x + 2) ( x^2 - 1)  =

 

(x + 2) ( x + 1) ( x - 1)

 

So.......the factored form of the polynomial is  ( x - 3)(x + 1) (x - 1) ( x + 2)

 

 

cool cool cool

 Sep 26, 2018

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