Question 1209877: Fill in the blanks, to make a true equation:
\frac{2x^4 - 3x^3 - x^2 + 4x - 4}{x^2 + x} = ___x^2 + ___x + ___ + ___/x + ___/(x + 1).
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's perform polynomial long division to find the quotient and remainder:
```
2x^2 - 5x - 3
____________________
x^2+x | 2x^4 - 3x^3 - x^2 + 4x - 4
2x^4 + 2x^3
--------------------
-5x^3 - x^2
-5x^3 - 5x^2
--------------------
4x^2 + 4x
4x^2 + 4x
--------------------
-4
```
So, we have:
$$ \frac{2x^4 - 3x^3 - x^2 + 4x - 4}{x^2 + x} = 2x^2 - 5x - 3 + \frac{-4}{x^2 + x} $$
Now, we need to perform partial fraction decomposition on $\frac{-4}{x^2 + x} = \frac{-4}{x(x + 1)}$.
We want to find constants A and B such that:
$$ \frac{-4}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} $$
Multiplying by $x(x + 1)$, we get:
$$ -4 = A(x + 1) + Bx $$
Let $x = 0$:
$$ -4 = A(1) + 0 \implies A = -4 $$
Let $x = -1$:
$$ -4 = 0 + B(-1) \implies B = 4 $$
So, we have:
$$ \frac{-4}{x(x + 1)} = \frac{-4}{x} + \frac{4}{x + 1} $$
Thus,
$$ \frac{2x^4 - 3x^3 - x^2 + 4x - 4}{x^2 + x} = 2x^2 - 5x - 3 + \frac{-4}{x} + \frac{4}{x + 1} $$
Therefore, the blanks are:
$$ 2x^2 - 5x - 3 + \frac{-4}{x} + \frac{4}{x + 1} $$
Final Answer: The final answer is $\boxed{2, -5, -3, -4, 4}$
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