Subjects algebra

Remainder Equality

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1. **State the problem:** Given the polynomial $$P(x) = (x^2+3)(x^2-1)$$ and two values $$m$$ and $$n$$ such that the remainders when $$P(x)$$ is divided by $$(x-m)$$ and $$(x-n)$$ are the same, show that $$m^3 + n^3 + m^2 n + m n^2 + 2m + 2n = 0$$. 2. **Understand the remainder condition:** When a polynomial $$P(x)$$ is divided by $$(x-a)$$, the remainder is $$P(a)$$ by the Remainder Theorem. 3. Since the remainders when dividing by $$(x-m)$$ and $$(x-n)$$ are the same, we have: $$P(m) = P(n)$$ 4. **Calculate $$P(x)$$ explicitly:** $$P(x) = (x^2 + 3)(x^2 - 1) = x^4 - x^2 + 3x^2 - 3 = x^4 + 2x^2 - 3$$ 5. **Evaluate $$P(m)$$ and $$P(n)$$:** $$P(m) = m^4 + 2m^2 - 3$$ $$P(n) = n^4 + 2n^2 - 3$$ 6. **Set the remainders equal:** $$m^4 + 2m^2 - 3 = n^4 + 2n^2 - 3$$ Simplify: $$m^4 + 2m^2 = n^4 + 2n^2$$ 7. **Rearrange:** $$m^4 - n^4 + 2(m^2 - n^2) = 0$$ 8. **Factor differences of powers:** $$m^4 - n^4 = (m^2 - n^2)(m^2 + n^2)$$ $$m^2 - n^2 = (m - n)(m + n)$$ So, $$(m^2 - n^2)(m^2 + n^2) + 2(m^2 - n^2) = (m^2 - n^2)(m^2 + n^2 + 2) = 0$$ 9. Since $$m eq n$$ (otherwise the problem is trivial), we have: $$(m^2 + n^2 + 2) = 0$$ 10. **Rewrite $$m^2 + n^2 + 2 = 0$$:** $$m^2 + n^2 = -2$$ 11. **Express $$m^2 + n^2$$ in terms of $$m+n$$ and $$mn$$:** $$m^2 + n^2 = (m+n)^2 - 2mn$$ So, $$(m+n)^2 - 2mn = -2$$ 12. **Let $$S = m+n$$ and $$P = mn$$, then:** $$S^2 - 2P = -2 \\ S^2 = 2P - 2$$ 13. **Now, consider the expression to prove:** $$m^3 + n^3 + m^2 n + m n^2 + 2m + 2n = 0$$ 14. **Group terms:** $$m^3 + n^3 + m^2 n + m n^2 + 2(m + n)$$ 15. **Rewrite $$m^3 + n^3$$ and $$m^2 n + m n^2$$:** $$m^3 + n^3 = (m + n)^3 - 3mn(m + n) = S^3 - 3PS$$ $$m^2 n + m n^2 = mn(m + n) = PS$$ 16. **Substitute back:** $$S^3 - 3PS + PS + 2S = S^3 - 2PS + 2S$$ 17. **Factor out $$S$$:** $$S^3 - 2PS + 2S = S(S^2 - 2P + 2)$$ 18. **Recall from step 12:** $$S^2 - 2P = -2$$ So, $$S^2 - 2P + 2 = -2 + 2 = 0$$ 19. **Therefore:** $$S(S^2 - 2P + 2) = S imes 0 = 0$$ 20. **Conclusion:** $$m^3 + n^3 + m^2 n + m n^2 + 2m + 2n = 0$$ is proven. **Final answer:** $$\boxed{m^3 + n^3 + m^2 n + m n^2 + 2m + 2n = 0}$$