Suppose p(x)=x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} and q(x)=x^{4}+b_{3}x^{3}+b_{2}x^{2}+b_{1}x+b_{0} are the polynomials. If \alpha, \beta, \gamma, \delta are zeros of p(x) and \alpha, \beta, \gamma, \lambda are zeros of q(x), then what is \frac{p(x)-q(x)}{(x-\alpha)(x-\beta)(x-\gamma)} equal to?

  1. A. -\lambda+\delta
  2. B. \lambda-\delta
  3. C. \lambda+\delta
  4. D. -\lambda-\delta

Correct Answer: B. \lambda-\delta

Explanation

We can write p(x) = (x-\alpha)(x-\beta)(x-\gamma)(x-\delta) and q(x) = (x-\alpha)(x-\beta)(x-\gamma)(x-\lambda). Subtracting q(x) from p(x) gives (x-\alpha)(x-\beta)(x-\gamma)( (x-\delta) - (x-\lambda) ) = (x-\alpha)(x-\beta)(x-\gamma)(\lambda - \delta). Dividing this by the common terms leaves \lambda-\delta.

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