Question

If x^2-1 is a factor of px^4 + qx^3 +rx^2 +sx+u,show that p+r+u=q+s=0.

Answer 1

X^2 -1=0
X^2 = 1
X = square root of 1
X = +1 OR -1
NOW AFTER GETTING THE VALUE OF "X" . WE INSERT THE VALUE IN THE OTHER EQUATION
PX^4 + QX^3 + RX^2 + SX + U = 0
WHEN X= +1 OR -1
P*-1^4 + Q*-1^3 + R*-1^2 + S*-1 + U = 0   ( WE TAKE X = -1)
P - Q + R - S + U = 0
THEREFORE,
P + R + U = Q + S
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