Answers:
a = 2
b = -1
c = 3
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Work Shown:
(numerator)/(denominator) = quotient + (remainder)/(denominator
(2x^3-5x^2+4x-8)/(ax^2+bx+c) = x-2 + (bx-a)/(ax^2+bx+c)
2x^3-5x^2+4x-8 = (x-2)(ax^2+bx+c) + (bx-a)
2x^3-5x^2+4x-8 = x(ax^2+bx+c)-2(ax^2+bx+c) + (bx-a)
2x^3-5x^2+4x-8 = ax^3+bx^2+cx-2ax^2-2bx-2c + bx-a
2x^3-5x^2+4x-8 = ax^3+(bx^2+2ax^2) + (-2bx+cx+bx) + (-a-2c)
2x^3-5x^2+4x-8 = ax^3+(b+2a)x^2 + (-b+c)x + (-a-2c)
The x^3 term of the left hand side (LHS) is 2x^3, while the x^3 term of the right hand side (RHS) is ax^3. This must mean a = 2.
The x^2 term of the LHS is -5x^2. The x^2 term of the RHS is (b+2a)x^2
Plugging in a = 2 leads to (b+2a)x^2 turning into (b+4)x^2. Equating the two x^2 coefficients has us get b+4 = -5 solve to b = -1
Now focus on the x terms. The LHS has 4x and the RHS has (-2b+c)x. Plug in b = -1 to get (-b+c)x turn into (c+1)x. Set this equal to 4x and you should find that c = 3
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In summary, we have:
a = 2
b = -1
c = 3
Note how the last term on the RHS is -a-2c = -2-2(3) = -8 which matches with the last term on the LHS. This helps confirm we have the correct values for a,b,c. Further confirmation would involve polynomial long division.
