## UnkleRhaukus roots of a trinomial $ax^3+bx^2+cx+d=0$ let the roots be $$p,q,r$$ When $$x$$ is equal a root the equation equals zero $(x-p)(x-q)(x-r)=0$ $(x-p)(x^2-(q+r)x+qr)=0$ $x\left(x^2-(q+r)x+qr\right)-p\left(x^2-(q+r)x+qr\right)=0$ $\left(x^3-(q+r)x^2+(qr)x\right)-\left(px^2-p(q+r)x-qpr\right)=0$ $x^3-(p+q+r)x^2+(qr+pq+pr)x-pqr=0$ comparing with $x^3+\frac bax^2+\frac cax+\frac da=0$ $-(p+q+r)=\frac ba\qquad\text{the sum of the roots is } -\frac ba$ $(qr+pq+pr)=\frac ca\qquad\text{sum of products of pairs of the roots is } \frac ca$ $-pqr=\frac da\qquad \text{the product of the roots is} -\frac da$ one year ago one year ago

1. mahmit2012

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2. mahmit2012

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3. mahmit2012

I divided to a.

4. UnkleRhaukus

5. mahmit2012

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6. UnkleRhaukus

i guess a trinomial always has at least one real root

7. mahmit2012

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8. mahmit2012

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9. hartnn

if the trinomial has real co-efficients, then complex roots occur in pairs, hence it must have atleast 1 real solution .

10. mahmit2012

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11. UnkleRhaukus

i see

12. mahmit2012

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13. mahmit2012

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14. mahmit2012

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15. mahmit2012

this is the general for all n-order polynomial equation.

16. UnkleRhaukus

awesome

17. hartnn

notice the sign change everytime, product of all roots = $$(-1)^n a_0/a_n$$

18. mahmit2012

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19. mahmit2012

I should mention that there are all real and complex roots in formulas.

20. mahmit2012

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21. mahmit2012

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