The polynomial f(x) =2x\(^3\) + px+ qx - 5 has (x-1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.
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Correct Answer: Option
Explanation:
f(x) =2xp\(^3\) + px\(^2\) + qx - 5
f(1) = 2(1)\(^3\) + p(1)\(^2\) + q(1) - 5
2+ p + q - 5;
p + q =3 -- (i)
f(-2) = 27
i.e, 2(-2)\(^3\) + p(-2)\(^2\) +q(-2) - 5 = 27
-16 +4p - 2q - 5 = 27
2p - q = 24--- (ii)
Solving(i) and (ii) Simultaneously
P = 9, q= -6
f(x) =2xp\(^3\) + px\(^2\) + qx - 5
f(1) = 2(1)\(^3\) + p(1)\(^2\) + q(1) - 5
2+ p + q - 5;
p + q =3 -- (i)
f(-2) = 27
i.e, 2(-2)\(^3\) + p(-2)\(^2\) +q(-2) - 5 = 27
-16 +4p - 2q - 5 = 27
2p - q = 24--- (ii)
Solving(i) and (ii) Simultaneously
P = 9, q= -6