If (8x3 - 27y3) ÷ (2x - 3y) = (Ax2 + Bxy + Cy2), then the value of (2A + B - C) is:
A. 4
B. 6
C. 5
D. 3
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \left( {8{x^3} - 27{y^3}} \right) \div \left( {2x - 3y} \right) = A{x^2} + Bxy + C{y^2} \cr & \frac{{\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right)}}{{\left( {2x - 3y} \right)}} = A{x^2} + Bxy + C{y^2} \cr & 4{x^2} + 6xy + 9{y^2} = A{x^2} + Bxy + C{y^2} \cr & {\text{Comparison both side}} \cr & A = 4,\,B = 6,\,C = 9 \cr & \left( {2A + B - C} \right) \cr & = \left( {2 \times 4 + 6 - 9} \right) \cr & = 5 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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