Question 52

If [x + p$$x^{3}$$ +q$$x^{2}$$ + rx + s] is divisible by ($$x^{2}$$ - 1), (x - 2) and (x - 3), values of p and s are respectively:

Solution

[x + p$$x^{3}$$ +q$$x^{2}$$ + rx + s] is divisible by ($$x^{2}$$ - 1), (x - 2) and (x - 3)

That means for x=1,-1,2,3 ,

we have four values of the equation:-

1. 1 + p + q + r + s =0

2. -1 -p + q -r + s = 0

3. 2 + 8p + 4q +2r +s = 0

4. 3 + 27p + 9q +3r +s = 0

Solving the four equations with four unknowns we get, p =-5, s=-6

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