# Find the value of

Let, f (x) = x4+px3+2x2-3x+q be the given polynomial.

And, let g (x) = (x2 – 1) = (x – 1) (x + 1)

Clearly,

(x – 1) and (x + 1) are factors of g (x)

Given, g (x) is a factor of f (x)

(x – 1) and (x + 1) are factors of f (x)

From factor theorem

If (x – 1) and (x + 1) are factors of f (x) then f (1) = 0 and f (-1) = 0 respectively.

f (1) = 0

(1)4 + p (1)3 + 2 (1)2 – 3 (1) + q = 0

1 + p + 2 – 3 + q = 0

p + q = 0 (i)

Similarly,

f (-1) = 0

(-1)4 + p (-1)3 + 2 (-1)2 - 3 (-1) + q = 0

1 – p + 2 + 3 + q = 0

q – p + 6 = 0 (ii)

Adding (i) and (ii), we get

p + q + q – p + 6 = 0

2q + 6 = 0

2q = - 6

q = -3

Putting value of q in (i), we get

p – 3 = 0

p = 3

Hence, x2 – 1 is divisible by f (x) when p = 3 and q = - 3.

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