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Question 81
Find the value of '$$m$$' if $$2x^m + x^3 - 3x^2 - 26$$ leaves a remainder of 994 when it is divided by $$x - 2$$.
Solution
$$2x^m + x^3 - 3x^2 - 26$$
$$Let f(x) = 2x^m + x^3 - 3x^2 - 26$$
When its divided with by (x-2), it leaves a remainder of f(2)
So f(2) = $$2\times(2)^m + 2^3 - 3\times2^2 - 26$$
=$$ 2^{m+1} + 8 - 12 - 26$$
=$$ 2^{m+1} - 30$$
which is equal to 994
$$ 2^{m+1} - 30 = 994$$
$$ 2^{m+1} = 1024$$
$$ 2^{m+1} = 2^{10}$$
$$m+1=10$$
$$m = 9$$
Option D is correct.
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