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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.


Solution

$$x^2-1=0\\\implies =(x+1)(x-1)=0\\\implies x=0,1$$
Now $$x^2-1$$ is a factor of $$px^4+qx^3+rx^2+sx+u$$
Putting $$x=1$$
we get,
$$p(1)^4+q(1)^3+r(1)^2+s(1)+u=0\\\implies p+q+r+s+u=0$$
Putting $$x=-1$$
we get,
$$p(-1)^4+q(-1)^3+r(-1)^2+s(-1)+u=0\\\implies p-q+r-s+u=0$$
Hence $$p+r+u=q+s=0$$


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