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Review the work showing the first few steps in writing a partial fraction decomposition.



StartFraction 2 x squared + 15 x + 19 Over (x + 4) cubed EndFraction = StartFraction A Over x + 4 EndFraction + StartFraction B Over (x + 4) squared + StartFraction C Over (x + 4) cubed EndFraction



2x2 + 15x + 19 = A(x + 4)2 + B(x + 4) + C



2x2 + 15x + 19 = Ax2 + 8Ax + 16A + Bx + 4B + C



Which system of equations can be used to determine the values of A, B, and C?

Sagot :

MrRoyal

Answer:

[tex]A = 2[/tex]

[tex]B = -1[/tex]

[tex]C = -9[/tex]

Step-by-step explanation:

Given

[tex]2x\² + 15x + 19 = Ax\² + 8Ax + 16A + Bx + 4B + C[/tex]

Required

Find A, B and C

Rearrange the expression

[tex]2x\² + 15x + 19 = Ax\² + 8Ax + 16A + Bx + 4B + C[/tex]

[tex]2x\² + 15x + 19 = Ax\² + 8Ax + Bx + 16A + 4B + C[/tex]

To do this, we simply compare the expressions on both sides of the equation.

So, we have:

[tex]Ax\² = 2x\²[/tex] --- (1)

[tex]8Ax + Bx = 15x[/tex] --- (2)

[tex]16A + 4B + C = 19[/tex] --- (3)

Divide both sides by x² in (1)

[tex]Ax\² = 2x\²[/tex]

[tex]A = 2[/tex]

Divide both sides by x in (2)

[tex]8Ax + Bx = 15x[/tex]

[tex]8A + B = 15[/tex]

Substitute 2 for A.

[tex]8*2 + B = 15[/tex]

[tex]16+ B = 15[/tex]

Make B the subject

[tex]B = 15 - 16[/tex]

[tex]B = -1[/tex]

Substitute 2 for A and -1 for B in (3)

[tex]16A + 4B + C = 19[/tex]

[tex]16*2 + 4*-1 + C = 19[/tex]

[tex]32 - 4 + C = 19[/tex]

[tex]28 + C = 19[/tex]

Make C the subject

[tex]C = 19 - 28[/tex]

[tex]C = -9[/tex]

Answer:

it D

Step-by-step explanation:

I just used elimination like a small brain

like so

when a = 1, b = 0, c = 3, d = -3

x^3 + 8x – 3 = Ax^3 + 5Ax + Bx^2 + 5B + Cx + D

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