+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?
+6251 \(\text{it's a pain showing division}\\ \text{forget about the }\pi \text{ for now}\\ \dfrac{x^3}{x}=x^2\\ x^3-5x^2-86x+360 - x^2(x-4) = \\ -x^2-86x+360\\ \dfrac{-x^2}{x}=-x\\ -x^2-86x+360 - (-x)(x-4) = \\ -90x+360\\ \dfrac{-90x}{x}=-90 -90x+360 -(-90)(x-4) = 0\)
\(\dfrac{x^3-5x^2-86x+360}{x-4} = x^2-x-90\)
\(\pi \dfrac{x^3-5x^2-86x+360}{x-4} = \pi (x^2-x-90)\)
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+38 Thanks, Rom, but I have to do it in stupid synthetic division or I get the entire question wrong.
+129852 Not that difficult, really...
4 [ 1pi -5pi -86pi 360pi ]
4pi -4pi -360pi
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1pi -pi -90pi 0
So....the remaining polynomial, p(x) , is
p(x) = pi [ x^2 - 1x - 90 ]
+38 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.
+129852 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
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1pi -1pi
Multiply 4 (-1pi) and put the result (-4pi)under the third term.....add again
4 [ 1 pi - 5pi - 86pi 360pi ]
4pi -4pi
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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
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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
+38 Okay, think I got it. Just double checking:
r | c1 c2 … cn
r•x1 … r•xn
x1 x2 … 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!
