If 2x3 + ax2 + bx - 2 leaves the remainders 7 and 0 when divided by (2x - 3) and (x + 2), respectively, then the values of a and b are respectively:
A. 3; -3
B. 2; -2
C. -3; 3
D. -2; 2
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & f\left( x \right) = 2{x^3} + a{x^2} + bx - 2 \cr & x + 2 = 0 \cr & x = - 2 \cr & f\left( { - 2} \right) = 2{\left( { - 2} \right)^3} + a{\left( { - 2} \right)^2} + b\left( { - 2} \right) - 2 = 0 \cr & \Rightarrow - 16 + 4a - 2b - 2 = 0 \cr & \Rightarrow 4a - 2b - 18 = 0 \cr & {\text{Now go through option - }} \cr & {\text{from option A}} \cr & 4 \times 3 - 2\left( { - 3} \right) - 18 = 0 \cr & 0 = 0{\text{ satisfy}} \cr} $$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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