Polynomial dividing

shahar

shahar

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I have a problem with two sub-problems:
1. Why the right side equal to the left side?
2. The same, but to show that the left side equal to the right side?

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shahar said:
I have a problem with two sub-problems:
1. Why the right side equal to the left side?
2. The same, but to show that the left side equal to the right side?

View attachment 14362
Click to expand...
Surely you know how to do long division on polynomials?
If not, take a refresher in baby algebra.
 
shahar said:
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1. When in doubt, multiply to run a quick check:
[math](x + 3) \left ( x^2 - 4x + 13 - \dfrac{40}{x + 3} \right ) = (x + 3) \cdot x^2 - (x + 3) \cdot -4x + (x + 3) \cdot 13 + (x + 3) \cdot \dfrac{-40}{x + 3} = x^3 - x^2 + x - 1[/math]
2. If the two are definite integrals over the same interval then they must be the same. Think about it in terms of "doing the same thing on both sides." (Like if y = 3x then we can say that [math]\left ( \dfrac{d}{dx} \right ) (y) = \left ( \dfrac{d}{dx} \right ) (3x)[/math] for derivatives.) If they are indefinite integrals then the best we can say is that the two sides are the same up to a constant.

-Dan
 
shahar said:
I have a problem with two sub-problems:
1. Why the right side equal to the left side?
2. The same, but to show that the left side equal to the right side?

View attachment 14362
Click to expand...
If you mean, (1) how can you obtain the RHS from the LHS, and (2) how can you obtain the LHS from the RHS, the answer is:
  1. Long division
  2. Add, using the common denominator x+3
 
topsquark said:
If they are indefinite integrals then the best we can say is that the two sides are the same up to a constant.
Click to expand...
Dan,
Am I missing something or if the two integrals are said to be equal then they do NOT differ by a constant? Shouldn't the constants be the same?
 
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