Given that (8x^(2)-6x+3)/((2x-1)^(2))=A+(B)/(2x-1)+(C)/((2x-1)^(2)), then A= _____, B= _____, C= _____

Set up equation: Set up the equation for partial fraction decomposition. We are given the equation: (8x^2 - 6x + 3) / ((2x - 1)^2) = A + (B / (2x - 1)) + (C / ((2x - 1)^2)) We need to find the values of A, B, and C.Clear fractions: Multiply both sides by the denominator to clear the fractions. Multiplying both sides by (2x - 1)^2, we get: 8x^2 - 6x + 3 = A(2x - 1)^2 + B(2x - 1) + CExpand right side: Expand the right side of the equation. Expanding A(2x - 1)^2 and B(2x - 1), we get: 8x^2 - 6x + 3 = A(4x^2 - 4x + 1) + B(2x - 1) + CEquate coefficients: Equate the coefficients of like terms on both sides of the equation. For the x^2 terms: 8 = 4A For the x terms: -6 = -4A + 2B For the constant terms: 3 = A - B + CSolve for A: Solve for A from the x^2 terms. Dividing both sides by 4, we get A = 2.Solve for B: Substitute A into the equation for the x terms and solve for B. -6 = -4(2) + 2B -6 = -8 + 2B 2B = 2 B = 1Solve for C: Substitute A and B into the equation for the constant terms and solve for C. 3 = 2 - 1 + C C = 3 - 2 + 1 C = 2

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Given that (8x^(2)-6x+3)/((2x-1)^(2))=A+(B)/(2x-1)+(C)/((2x-1)^(2)), then
A= _____,
B= _____,
C= _____

Given that $\frac{8 x^{2}-6 x+3}{(2 x-1)^{2}}=A+\frac{B}{2 x-1}+\frac{C}{(2 x-1)^{2}}$ , then A= ___, B= _, C= ___

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Algebra 1

Transformations of quadratic functions

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Q. Given that

\frac{8 x^{2}-6 x+3}{(2 x-1)^{2}}=A+\frac{B}{2 x-1}+\frac{C}{(2 x-1)^{2}}

, then A= _____, B= _____, C= _____

Set up equation: Set up the equation for partial fraction decomposition. $\newline$ We are given the equation: $\newline$ $(8x^2 - 6x + 3) / ((2x - 1)^2) = A + (B / (2x - 1)) + (C / ((2x - 1)^2))$ $\newline$ We need to find the values of $A$ , $B$ , and $C$ .
Clear fractions: Multiply both sides by the denominator to clear the fractions. $\newline$ Multiplying both sides by $(2x - 1)^2$ , we get: $\newline$ $8x^2 - 6x + 3 = A(2x - 1)^2 + B(2x - 1) + C$
Expand right side: Expand the right side of the equation. $\newline$ Expanding $A(2x - 1)^2$ and $B(2x - 1)$ , we get: $\newline$ $8x^2 - 6x + 3 = A(4x^2 - 4x + 1) + B(2x - 1) + C$
Equate coefficients: Equate the coefficients of like terms on both sides of the equation. $\newline$ For the $x^2$ terms: $8 = 4A$ $\newline$ For the $x$ terms: $-6 = -4A + 2B$ $\newline$ For the constant terms: $3 = A - B + C$
Solve for $A$ : Solve for $A$ from the $x^2$ terms. $\newline$ Dividing both sides by $4$ , we get $A = 2$ .
Solve for B: Substitute $A$ into the equation for the $x$ terms and solve for $B$ . $\newline$ $-6 = -4(2) + 2B$ $\newline$ $-6 = -8 + 2B$ $\newline$ $2B = 2$ $\newline$ $B = 1$
Solve for C: Substitute $A$ and $B$ into the equation for the constant terms and solve for $C$ . $3 = 2 - 1 + C$ $C = 3 - 2 + 1$ $C = 2$