+4609 \(\frac{(80x^3-50x^2+7)}{(8x-5)}\). 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.
+2448 That's definitely the right answer but not how it's supposed to be solved, but thanks! I think I got it ^-^
+128474 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????
+2448 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
+128474 Different roads....same answer....good job !!!
[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 !!! ]
