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# Help with Synthetic Division

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

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

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

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Feb 10, 2019
#4
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Thanks, Rom, but I have to do it in stupid synthetic division or I get the entire question wrong.

#2
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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 ]   Feb 10, 2019
#3
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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.  #5
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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

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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   Feb 10, 2019
#6
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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
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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   CPhill  Feb 10, 2019
edited by CPhill  Feb 10, 2019