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(2x^(2)+ 7x -4 )/(x-2 )=ax + b + (c)/(x-2)

find a , b and c
thank you so much i have an exam after a few days
help me
ty again

User Scutnex
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1 Answer

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To find a, b, and c, we can apply the method of partial fractions to the given expression.

First, we write the expression as:

(2x^2 + 7x - 4)/(x - 2) = [(ax + b) + c/(x - 2)]/(x - 2)

Then, we multiply both sides of the equation by (x - 2) to get:

2x^2 + 7x - 4 = (ax + b)(x - 2) + c

Expanding the right side gives:

2x^2 + 7x - 4 = ax^2 - 2abx + 2b + c

Matching coefficients on both sides, we get the following system of equations:

a + 0 + 0 = 2

0 - 2b + 0 = 7

0 + 0 + 1 = -4

Solving this system, we find that a = 2, b = -3.5, and c = 1.

Therefore, the values of a, b, and c in the expression (2x^2 + 7x - 4)/(x - 2) = ax + b + c/(x - 2) are a = 2, b = -3.5, and c = 1.
User Rom Eh
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