If 8x2−12x(x2+2x−6)=Ax+Bx+Cx2+2x−6 then (A,C)=

8x2−12=A(x2+2x−6)+(Bx+C)xPut x=0⇒−12=−6A⇒A=2∴(A,C)=(2,−4)

If 8x2−12x(x2+2x−6)=Ax+Bx+Cx2+2x−6 then (A,C)=

8x2−12=A(x2+2x−6)+(Bx+C)xPut x=0⇒−12=−6A⇒A=2∴(A,C)=(2,−4)

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If $\frac{8 x^{2} - 12}{x (x^{2} + 2 x - 6)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 2 x - 6}$ then $(A, C) =$

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$(2, 4)$

$(- 2, - 4)$

$(- 2, 4)$

$(2, - 4)$

answer is D.

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Detailed Solution

$8 x^{2} - 12 = A (x^{2} + 2 x - 6) + (B x + C) x$

Put $x = 0 \Rightarrow - 12 = - 6 A \Rightarrow A = 2$

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$∴ (A, C) = (2, - 4)$

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If 8x2−12x(x2+2x−6)=Ax+Bx+Cx2+2x−6 then (A,C)=

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If $\frac{8 x^{2} - 12}{x (x^{2} + 2 x - 6)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 2 x - 6}$ then $(A, C) =$