CBSE Class 9 Answered

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 x = 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 b = – 239, c = 150 and k = – 90

Hence, the value of k is – 90.