+0  
 
+1
54
7
avatar+38 

I have to show that x-4 is a factor of v(x)= pi(x3-5x2-86x+360) using synthetic division and I have to show my work, but I don’t understand synthetic division. Or factoring equations. Could someone help me, please?

 Feb 10, 2019
 #1
avatar+4404 
+1

\(\text{it's a pain showing division}\\ \text{forget about the }\pi \text{ for now}\\ \dfrac{x^3}{x}=x^2\\ x^3-5x^2-86x+360 - x^2(x-4) = \\ -x^2-86x+360\\ \dfrac{-x^2}{x}=-x\\ -x^2-86x+360 - (-x)(x-4) = \\ -90x+360\\ \dfrac{-90x}{x}=-90 -90x+360 -(-90)(x-4) = 0\)

 

\(\dfrac{x^3-5x^2-86x+360}{x-4} = x^2-x-90\)

 

\(\pi \dfrac{x^3-5x^2-86x+360}{x-4} = \pi (x^2-x-90)\)

.
 Feb 10, 2019
 #4
avatar+38 
0

Thanks, Rom, but I have to do it in stupid synthetic division or I get the entire question wrong.

IeweadOTuell  Feb 10, 2019
 #2
avatar+98060 
+2

Not that difficult, really...

 

4 [    1pi          -5pi          -86pi          360pi     ]

                        4pi           -4pi          -360pi                 

       ___________________________________

        1pi          -pi           -90pi             0

 

 

So....the remaining polynomial, p(x) ,  is  

 

p(x)  = pi [ x^2 - 1x   - 90 ]

 

 

 

cool cool cool

 Feb 10, 2019
 #3
avatar+38 
+2

Thanks, CPhill! So, essentially, when you use synthetic division, remove the variable, execute long division on each coefficient by the factor you are testing, and reinsert the variable with respect to each part's powers? Or am I still wrong? Because I'm still quite lost with synthetic division. I mean, I would use long division, but no, we have to use synthetic division because it makes that much of a difference in the long term.

angrycheeky

IeweadOTuell  Feb 10, 2019
 #5
avatar+98060 
+2

It's really pretty simple

 

 

Since we   are   testing if x - 4  is a factor....we actually want to see if 4 is a root....this is what we divide by

 

The coefficients (and constant term ) of the polynomial are then written.....be careful....any missing power is represented by 0

 

So we have

 

4 [    1pi          -5pi          -86pi          360pi     ]              bring down the first term

 

      ___________________________________  

        1pi                                                                

 

Multiply this by the divisor....and put this result  (4pi) under the second term....then add

So we have

 

4 [ 1 pi     - 5pi       - 86pi       360pi   ]

                  4pi

     ____________________________       

     1pi       -1pi   

 

Multiply  4 (-1pi)  and put the result  (-4pi)under the third term.....add again 

 

4 [ 1 pi     - 5pi       - 86pi       360pi   ]

                  4pi        -4pi      

     ____________________________       

     1pi       -1pi      - 90pi 

 

Multiply  4(-90i)  and put the result  (-360 pi) under the last term....add again and we get 0

 

   

4 [ 1 pi     - 5pi       - 86pi       360pi   ]

                  4pi        -4pi        -360pi      

     ____________________________       

     1pi       -1pi      - 90pi          0

 

The remaining polynomial  is  one degree less than the one we started with   ..so...    degree 2

 

So we have

 

1pi x^2   -  pi x  - 90 pi           factoring out pi, we get

 

pi [ x^2 - x  - 90 ]   =  the remaining polynomial

 

And that's it!!!

 

Hope that helped

 

 

cool cool cool

 Feb 10, 2019
 #6
avatar+38 
+1

Okay, think I got it. Just double checking:

 

     |     c1       c2   …   cn

                           r•x1  …  r•xn

                x1         x  …   xn

Did I get it? (Pls say yes, pls say yes...) XD Thanks a ton man! You probably save my grade from going to an F!

 Feb 10, 2019
 #7
avatar+98060 
+1

r [  c1     c2                      c3                         c4     ....                          ]

           r*c1                 r * [  r*c1 + c2]      r [  r * [ r*c1 + c2] + c3 ]        ]

     ____________________________________________________

     c1    r*c1 + c2     r * [ r*c1 + c2] + c3     r [  [ r*c1 + c2] + c3 ] + c4       .....etc

 

 

cool cool cool

CPhill  Feb 10, 2019
edited by CPhill  Feb 10, 2019

26 Online Users

avatar