How can I find the polynomial equation with the roots -1, (2+2√ 5)/2, (2-2√ 5)/2.

Question

anonymous · Answer

$$\frac{ 2+2\sqrt{5} }{ 2 }$$ $$\frac{ 2-2\sqrt{5} }{ 2 }$$

anonymous · Answer

Since there are 3 roots, the equation should be third degree.
y = (x + 1)(ax^2 + bx + c ) = 0. Find a, b, and c
Product of the 2 real roots: c/a = (1 + V5)(1 - V5) = 1 - 5 = -4 -> c = -4a (1).
Root x1 = -b/2a + VD/2a = (2 + V5)/2 -> a = 1 and b = -2
c = -4a = -4. Finally: y = (x + 1)(x^2 - 2x - 4)
Check: D = b^2 - 4ac = 4 + 16 = 20 = 4*5--> VD = 2V5
Roots: x1 = 2/2 + 2V5/2. Correct.

anonymous · Answer

That's absolutely correct, yes. Could you perhaps explain how you found c a little more though? I want to learn how to do it myself as opposed to copy and paste your answer because this has completely stumped me.

anonymous · Answer

Such as why you did the (1+V5)(1-V5) and what your logic was for the subsequent 5 in the 1-5.

anonymous · Answer

The product of the 2 real roots is (c/a).
The roots from the text are, after simplification: (1 + V5) and (1 - V5).
Product: (1 + V5)(1 - V5) = 1 - 5 = -4   (identity (a - b)( a = b) = a^2 - b^2)
Since a/c = -4, then c = -4a.