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# ​ Synthetic division

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225
8 Another one >.<

would it be 5/8 for 8x-5

Sep 19, 2018
edited by RainbowPanda  Sep 19, 2018

#1
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$$\frac{(80x^3-50x^2+7)}{(8x-5)}$$. First, divide the leading coefficients of $$80x^3-50x^2+7\mathrm{\:and\:the\:divisor\:}8x-5$$, so this means that:

$$\frac{80x^3}{8x}$$, which is $$10x^2.$$ That is our quotient, and now we have to find the remainder. So, multiply 8x-5 by 10x^2, to attain $$80x^3-50x^2.$$ Now, subtract this result from $$80x^3-50x^2+7$$ , to get $$7$$ as our new remainder. Therefore, $$\frac{(80x^3-50x^2+7)}{(8x-5)}=\boxed{10x^2+\frac{7}{8x-5}}$$.

Wait, is this good? IDK.  Sep 19, 2018
#2
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That's definitely the right answer but not how it's supposed to be solved, but thanks! I think I got it ^-^

RainbowPanda  Sep 19, 2018
#3
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Yeah, I didn't use synthetic division.

tertre  Sep 19, 2018
#5
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That's OK, tertre.....synthetic division is a little bit tricky in this kind of problem when the linear divisor  is in the form ax +/- b    !!!   CPhill  Sep 19, 2018
#4
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Set 8x  - 5  = 0  ⇒    x =  5/8....this is what we need to divide by

Note that the polynomial is really   80x^3  - 50x^2  + 0x   + 7

5/8  [  80      -50       0        7  ]

50       0        0

_____________________

80         0        0         7

The remainder is correct

The apparent remaining polynomial is  80x^2

Note....RP...that all we really need to do to find the correct remaining polynomial is just to divide  the apparent remaining polynomial  by   the "a"  coefiicient  of the  divisor, ax - b.....in this case.....ax - b  = 8x  - 5...so we can divide

80x^2  by   8  = 10x^2  and this is the correct residual polynomial

So...the answer   10x^2 R [ 7 /(8x - 5) ]

Does that help????   Sep 19, 2018
edited by CPhill  Sep 19, 2018
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Yes I did it a different way if that's alright. 5/8| 10   -25/4   0   7/8

25/4   0     0

-------------------------

10   0     0     7/8

10x^2+7/8x-5

RainbowPanda  Sep 19, 2018
#7
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[You divided the polynomial by 8 right away.....and realized that the  "7"  in the denominator of the "remainder fraction" represented the true remainder....if you like that method better, stick with it...I just learned it a different way....at least you understand these, now....that's the important thing  !!! ]   CPhill  Sep 19, 2018
#8
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Wow! Great Solution, CPhill! tertre  Sep 19, 2018