Q. 3 B5.0( 1 Vote )

Given p(x) = x3 + x2 + x

Let the number to be added be “k”.

Then, the new polynomial q(x) = x3 + x2 + x + k

Now, (x+1) is a factor of x3 + x2 + x + k.

i.e. x = – 1 is the root of the polynomial.

Then, put the polynomial to zero we get,

x3 + x2 + x + k = 0

( – 1)3 + ( – 1)2 + ( – 1) + k = 0

– 1 + 1 – 1 + k = 0

k = – 1

Hence, “ – 1” should be added to the polynomial such that (x – 1) is a factor of q(x).

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