# Find the value of p and q at the polynomial x(power4) + p x(x power 3) + 2x(power 2)- 3x + q is divisible by (x-1) and (x+1).

2
by hrithi

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by hrithi

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Here, x-1 and x + 1 are its factors.

so, x - 1 = 0, x + 1 = 0

x = 1, x= -1

p(1) = 0

p(-1) = 0

p(1) =

0 = 1 + p + 2 - 3 + q

0 = p +q

-p = q eqn 1

p(-1) =

0 = 1 - p +2 +3 +q

0 = 6 - p + q

from eqn 1 -p = q

0 = 6 + q + q

6 = 2q

3 = q

-p = q

p = -q

p = -3

and (x-1) & (x+1)

x-1 = 0

therfore x = 1

X+1 = 0

x = -1

therfore, x = (1,-1)

Substituting x= 1 in (1)

therfore

1⁴+px³+2x²-3x+q = 0

1 + p + 2 - 3 +q =0

p+q = 0 -----------------(2)

Substituting x=(-1)

(-1)⁴+p(-1)³+2(-1)²-3(-1)+q = 0

1 -p+ 2 + 3 + q =0

-p+q = --6---------------------(3)

(2) + (3) =

2q= --6

q = -3

substituting q = -3 in (2)

p+q=0

p+(-3)=0

p=3

q=-3