SOLUTION: long division: (x^3-3x^2+2x-8)/(x^2+4)
and find the 9th term of:
(x^2-2)^9
I need help i'm completely confused on long division and terms. I don't understand the formulas.
Algebra ->
Sequences-and-series
-> SOLUTION: long division: (x^3-3x^2+2x-8)/(x^2+4)
and find the 9th term of:
(x^2-2)^9
I need help i'm completely confused on long division and terms. I don't understand the formulas.
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Question 1060705: long division: (x^3-3x^2+2x-8)/(x^2+4)
and find the 9th term of:
(x^2-2)^9
I need help i'm completely confused on long division and terms. I don't understand the formulas. Can you help? Found 2 solutions by josgarithmetic, MathTherapy:Answer by josgarithmetic(39618) (Show Source):
You can put this solution on YOUR website! The second question is reliant on simply following the Binomial Theorem. You can refer to that and fill-in what you need.
The long division for the polynomials works the same way as long division for regular base-ten numbers, and can be done without the condensing that you would be able if in base-ten numbers.
This could be broken into steps with explanations, but here I will just show the whole thing in one finished piece:
The dividend is shown toward the right, and the divisor is shown at the left, and you begin first asking, what is x^3 divided by x^2; and this is x; and so you put in the first partial quotient of x, ABOVE the x^3.
You can put this solution on YOUR website! long division: (x^3-3x^2+2x-8)/(x^2+4)
and find the 9th term of:
(x^2-2)^9
I need help i'm completely confused on long division and terms. I don't understand the formulas. Can you help?
The formula you need for the 9th term of a BINOMIAL EXPANSION is: , where, in this case:
a =
b = - 2
n = 9
r, or term number = 9
Therefore, becomes:
It is that easy....nothing too COMPLEX!