the remainder

Asked by manvendra singh | 19th Nov, 2013, 10:39: PM

Expert Answer:

Let p(x) = x 5 + kx 2

Consider division of (x 5 + kx 2) by (x – 1) (x – 2) (x – 3), then by division algorithm, there exists polynomials Q(x) and R(x) such that

x 5 + kx 2 = (x-1)(x-2)(x-3)Q(x) + R(x)

Here, 0 less than equal to deg R(x) < deg (x-1)(x-2)(x-3)

Let R(x) = ax2+bx+c

But it is given that remainder does not contain any therm of x 2

So, a = 0

This gives R(x) = bx+c

Therefore,

x 5 + kx 2 = (x-1)(x-2)(x-3)Q(x) + (bx+c)          ... (1)

On putting = 1, 2 and 3 in (1), we get

1 + k = b + c, 32 + 4k = 2b + c and 243 + 9k = 3b + c

On solving the equations, we get = – 239, c = 150 and k = – 90

Hence, the value of k is – 90.

Answered by  | 19th Nov, 2013, 11:19: PM

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