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3 years ago

Using the polynomials Q = 3 x2 + 5 x - 2, R = 2 - x2 , and S = 2 x + 5, perform the indicated operation.

Mathematics

1 answer:

marusya05 [52]3 years ago

4 0

Answer:

4x^2 + 3x - 9

Step-by-step explanation:

First, add together functions R(x) and S(x). Stack them vertically and combine like terms:

R(x) = 2 - x^2

S(x) = 2x + 5

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

(R+S)(x) = -x^2 + 2x + 7

Now subtract the above result from Q(x):

Q(x) - (R(x) + S(x)) = 3x^2 + 5x - 2 + x^2 - 2x - 7, or 4x^2 + 3x - 9

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(6x - 5y + 4) dy + (y - 2x - 1) dx = 0

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Let X = x - a and Y = y - b. We want to find constants a and b such that

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where the numerator and denominator on the right side are free of constant terms. Substituting x and y in the equation, we have

dY/dX = (2 (X + a) - (Y + b) + 1) / (6 (X + a) - 5 (Y + b) + 4)

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Then we solve for a and b in the system,

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With these constants, the equation reduces to

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X dV/dX + V = (2X - VX) / (6X - 5VX)

The equation becomes separable after some simplification:

X dV/dX + V = (2 - V) / (6 - 5V)

X dV/dX = (2 - V) / (6 - 5V) - V

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Integrate both sides:

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Extract a constant from the logarithm on the left:

-5/4 V + 1/4 (ln(4) + ln|V - 1|) = ln|X| + C

-5/4 V + 1/4 ln|V - 1| = ln|X| + C

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Get this back in terms of Y :

-5Y/X + ln|Y/X - 1| = 4 ln|X| + C

Now get the solution in terms of y and x :

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With some manipulation of constants and logarithms, and a bit of algebra, we can rewrite this solution as

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