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A, B and C are constants such that for all values of x, 3x2 + 6x + 5 = (Ax + B)(x - 3) + C(x2 + 1). Find the value of A.
  • a)
    0
  • b)
    -2
  • c)
    -3
  • d)
    -5
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A, B and C are constants such that for all values of x, 3x2 + 6x + 5 ...
To find the value of A, we need to equate the coefficients of x on both sides of the equation.

The given equation is: 3x^2 + 6x + 5 = (Ax + B)(x - 3) + C(x^2 + 1)

Expanding the right side of the equation, we get: 3x^2 + 6x + 5 = (Ax^2 - 3Ax + Bx - 3B) + Cx^2 + C

Now, let's equate the coefficients of x^2 on both sides:

On the left side, the coefficient of x^2 is 3.
On the right side, the coefficient of x^2 is A + C.

Therefore, we have the equation: A + C = 3 ----(1)

Next, let's equate the coefficients of x on both sides:

On the left side, the coefficient of x is 6.
On the right side, the coefficient of x is -3A + B + C.

Therefore, we have the equation: -3A + B + C = 6 ----(2)

We also know that the constant term on both sides should be equal:

On the left side, the constant term is 5.
On the right side, the constant term is -3B + C.

Therefore, we have the equation: -3B + C = 5 ----(3)

Now, we have a system of three equations (1), (2), and (3) with three variables A, B, and C. We can solve this system to find the values of A, B, and C.

Adding equations (1) and (3), we get: A - 3B + 2C = 8 ----(4)

Subtracting equation (3) from equation (2), we get: -3A + B = 1 ----(5)

Now, we can solve equations (4) and (5) simultaneously to find the values of A and B.

Multiplying equation (5) by 3 and adding it to equation (4), we get: 10C = 11

Simplifying, we find: C = 11/10

Substituting the value of C in equation (1), we get: A + 11/10 = 3

Simplifying, we find: A = 19/10

Therefore, the value of A is 19/10, which is approximately equal to -1.9.

The correct answer is option (b) -2.
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Community Answer
A, B and C are constants such that for all values of x, 3x2 + 6x + 5 ...
3x2 + 6x + 5 = Ax2 - 3Ax + Bx - 3B + Cx2 + C
3x2 + 6x + 5 = (A + C)x2 + (B - 3A)x + (C - 3B)
A + C = 3 .... (1)
B - 3A = 6 .... (2)
C - 3B = 5 ... (3)
B - 3(3 - C) = 6 ... (4)
Or B - 9 + 3C = 6
From (3),
C = 5 + 3B
B - 9 + 3(5 + 3B) = 6
B - 9 + 15 + 9B = 6
0B = 0 B = 0
Or, C = 5
Or, A = 3 - 5 = -2
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